207 mt. douglas secondary how to use ti-83, ti …...how to use ti-83, ti-83 plus, ti-84 plus...

Foundations of Math 11 Calculator Usage ♦ 207

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HOW TO USE TI-83, TI-83 PLUS, TI-84 PLUS CALCULATORS

FOR STATISTICS CALCULATIONS

shows it is an actual calculator key to press

1. Using LISTS to Calculate Mean, Median and Standard Deviation

A. To clear existing values from L1 and L2:

STAT (move cursor down to ClrList) ENTER

2nd L1 , 2nd L2 ENTER

ClrList L1 , L2 Done

B. To enter new values in L1 AND L2:

STAT (Edit) ENTER

(now use cursor key to enter data values in L1 and frequencies in L2)

L1 L2 85 2 95 9 105 11 � �

C. To calculate mean, median and standard deviation:

STAT (move cursor to CALC) ENTER


1 – Var Stats L1 L2

D. Read mean → x =

Standard deviation → σ x =

(cursor down 3 times) Median → Med =

1 – Var Stats

x =

�

σ x =

� Med =

2. To Calculate the Area under a Normal Curve

A. Method 1 using normalcdf

2nd DISTR (move cursor to 2: normalcdf) ENTER

Syntax: normalcdf (Za ,Zb ) → area between Za and Zb

normalcdf (a, b, μ, σ) → area between a and b use – 1E99 for left extreme (or –99) and 1E99 for right extreme (or 99)

e.g., normalcdf ( – 1E99, b, μ, σ) → area to left of b normalcdf (a, 1E99, μ, σ) → area to right of a

B. Method 2 using ShadeNorm ( Za , Zb )

i) Set WINDOW values:

Xmin = −3 ← Xmin = μ − 3σ = 0− 3×1

Xmax = 3 ← Xmax = μ + 3σ = 0+ 3×1

Ymax = 0.4 ← Xmax = 0.4 ÷ σ = 0.4 ÷1

Ymin = −0.1 ← Ymin = −(Ymax ÷ 4) allows room for text

Mt. Douglas Secondary

208 ♦ Calculator Usage Foundations of Math 11


ii) Clear previous normal curve:

2nd DRAW (ClrDraw) ENTER ENTER

ClrDraw Done

iii) Draw the normal curve:

2nd DISTR (move cursor to DRAW ) ENTER

ShadeNorm ( Za , Zb )

,

iv) Once you have finished recording results, ClrDraw again:

See step (ii) above.

ClrDraw Done

C. Method 3 using ShadeNorm (a, b, μ , σ) Example: find the probability that a person has an IQ < 120 given the

mean IQ is 100, with standard derviation 15.

i) CALCULATE WINDOW VALUES, (using μ and σ)

μ = 100, σ = 15

Xmin = μ – 3σ = 100 – 3 × 15 = 55

Xmax = μ + 3σ = 100 + 3 × 15 = 145

Xmax = 0.4 ÷ σ = 0.4 ÷ 15 ≈ 0.03

Ymin = −(Ymax ÷ 4) = −(0.03) ÷ 4 ≈ – 0.01

ii) Clear previous normal curve:

See step (ii) in Method II.

ClrDraw Done

iii) Draw the normal curve:

2nd DISTR DRAW ENTER

ShadeNorm (– 1E99, 120, 100, 15) ENTER

,

Syntax: ShadeNorm (a, b, μ, σ)

use a = – 1E99 for left extreme → area to left of b

use b = 1E99 for right extreme → area to right of a

Note: You can also use –99 and 99 for the left and right extremes.


Foundations of Math 11 Calculator Usage ♦ 209


3. To Calculate Z-score for a Standard Normal Distribution

A. Method 1 Using invNorm (area)

x − uσ

= Z → Z = invNorm (area)

2nd DISTR (move cursor to) invNorm ENTER (enter value) ENTER

Note: the area is for a Normal Curve.

B. Method 2 Using invNorm (area, μ, σ )

2nd DISTR (move cursor to) invNorm ENTER (enter values) ENTER

Enter z-score area, then μ = mean , σ = standard deviation

Note: the area is for a Standard Normal Curve.

Note: area is zero at – 1E99 and one at 1E99, therefore, area varies from left to right between

zero and one.


210 ♦ Chapter 5 – Statistics Foundations of Math 11



Foundations of Math 11 Section 5.1 – Mean, Median and Mode ♦ 211


5.1 Mean, Median and Mode

Statistics is a field of Mathematics dealing with the collecting and summarizing of data. Once the data has been gathered, the information is evaluated and analyzed so that a decision based on these measurable events can be made. Society in general depends to a great extent on our ability to evaluate information, determine what is true, and make correct decisions. It is amusing to remember the famous words of Mark Twain when interpreting information, “There are three kinds of lies — lies, damn lies, and statistics.” Mean, Median and Mode

Mean, median and mode are all measures of central tendency or average. These represent, respectively, “average,” “centre” and “most frequent” of the data gathered. Mean — The mean is computed by adding a set of values and dividing by the total number of values.

x = mean of a sample, μ = mean of the whole population.

i) μ =

x1 + x2 + x3 +…+ xn

n • basic mean formula

ii) μ =

xii=1

n

∑n

• mean using sigma notation

iii) μ =

fixii=1

n

∑n

• mean using group distribution

fi = frequency

Note: the sum of n numbers with a mean of μ is Sn = n μ .

Example 1 Determine the mean of: 1, 6, 3, 8, 9, 3

Solution:▼ μ = 1+ 6 + 3+ 8+ 9 + 3

6 = 5




Example 2 For 30 randomly selected high school students, the following IQ frequency distribution was obtained. Determine the mean.

Class Limits Frequency

80 ≤ x < 90 2

90 ≤ x < 100 9

100 ≤ x < 110 11

110 ≤ x < 120 5

120 ≤ x < 130 2

130 ≤ x < 140 1

Solution:▼ Method 1

Midpoint of class limits are 85, 95, 105, 115, 125, 135

x = 2 × 85+ 9 × 95+11×105+ 5×115+ 2 ×125+1×135

30= 104.7

Method 2 (by TI-83 Calculator)

STAT (EDIT) ENTER

(enter values under L1 & L2) STAT (move cursor to CALC) ENTER (1–Var

Stats)


Example 3 10 numbers have a mean of 37. If one number is removed, the mean is 38. What was the number that was removed?

Solution:▼ Sum of 10 numbers: S10 = 10 × 37 = 370

Sum of 9 numbers: S9 = 9 × 38 = 342

Removed number was: 370 – 342 = 28

Mean




Median — The median is the middle value in an odd number of values. In an even number of values, the

median is the mean of the two middle values. Before one can find this point, the data must be arranged in order.

By formula, the median is the n +1

2 term.

Example 1 Odd number of values:

7

5

4 ← Median = 4

2

1

Solution:▼ 5 terms, so the median is 5+1

2= 3rd term

Example 2 Even number of values:

2

3

4

7 }← Median =

4 + 7

2 = 5.5

7

8

Solution:▼ 6 terms, so the median is 6 +1

2= 3.5 , therefore, median =

3rd term + 4th term

2

Example 3 For 30 randomly selected high school students, the following IQ frequency distribution was obtained. Determine the median.


80 ≤ x < 90 2

90 ≤ x < 100 9

100 ≤ x < 110 11

110 ≤ x < 120 5

120 ≤ x < 130 2

130 ≤ x < 140 1




Median


By formula, the median term is n +1

2= 30 +1

2= 15.5

So, the median is 15th term + 16th term

2.

Both the 15th term and 16th term are found in the interval 100 ≤ x < 110 so the median is 105.


STAT (EDIT) ENTER

(enter values under L1 & L2) STAT (move cursor to CALC) ENTER

(1–Var Stats) 2nd L1 , 2nd

L2 ENTER (move cursor down to Med)

Mode — The mode is the value that occurs most often in a set of values. If all values are equal in frequency,

then there is no mode; there can be multiple modes if same values occur equally often.

Example 1 What is the mode of the following data set?

0, 0, 1, 1, 1, 2, 2, 2, 3, 3

Solution:▼ Both 1 and 2 have a frequency of 3, therefore the modes are 1 and 2.

Example 2 For 30 randomly selected high school students, the following IQ frequency distribution was obtained. Determine the mode.


80 ≤ x < 90 2

90 ≤ x < 100 9

100 ≤ x < 110 11

110 ≤ x < 120 5

120 ≤ x < 130 2

130 ≤ x < 140 1

Solution:▼ The range 100 ≤ x <110 has frequency of 11, so the mode is 105.




5.1 Exercise Set

1. The value of the middle term in a ranked data

set is called the

a) mean b) median c) mode

2. Which of the following is affected by

extreme values?


3. Which of the following can have more than

one value?


4. Determine the mean, median and mode for the

following set of values:

1, 2, 3, 4, 4, 7

5. The incomes of a sample of 6 local teachers

are as follows: $41 500, $44 900, $39 700,

$62 300, $58 500 and $53 100. What is

the mean, median and mode income of the

6 teachers?

6. Determine the mean, median and mode

salaries of the staff listed below:

Staff Salary

One owner $80 000

One Manager $60 000

Two salespersons $48 000

Six technicians $44 000




7. The following frequency distributions

represents the monthly commission in dollars

for 25 car salespersons at a car lot. Determine

the mean, median and mode.

Commission in $ Frequency

800 ≤ x < 1600 3

1600 ≤ x < 2400 4

2400 ≤ x < 3200 6

3200 ≤ x < 4000 12

8. The following table gives the frequency

distribution of the number of orders received

each day during the past 50 days at the office

of a publishing company.

Number of Orders Number of Days

10–12 7

13–15 12

16–18 17

19–21 14

Calculate the mean, median and mode.

9. The mean age of five people is 39. The ages

of four of these five persons are 33, 45, 27 and

41. Find the age of the fifth person.

10. The mean score of 18 female students on

a math test is 72 and the mean score of

14 male students is 66. Find the combined

mean score.




11. A small business has 10 people in total on

the payroll.

– 8 workers make $10 000 per year

– 1 foreman makes $40 000 per year

– 1 owner makes $880 000 per year

a) Determine the mean, median and mode for the 10 people.

b) If you were the owner, what type of average would you prefer to use in wage bargaining? Why?

c) If you were a worker, what type of average would you prefer to use in wage bargaining? Why?

12. a) What form of central tendency is always part of the data?

b) What form of central tendency requires using all values of the data?

c) What form of central tendency is the centre or middle value of the data?

d) What form of central tendency is least likely to be an actual score of the data?

e) Which of the three values of central tendency can assume more than one value for the data?

13. Consider the data sets

Set I: 5 9 16 10 11

Set II: 11 15 22 16 17

Notice that each value of the second data set is obtained by adding 6 to the corresponding value of the first data set. Calculate the mean of the two data sets, and comment on the relationship between the two means.

14. Consider the data sets

Set I: 5 9 16 10 11

Set II: 10 18 32 20 22

Notice that each value of data set II is obtained by multiplying the corresponding value of the first data set by 2. Calculate the mean of these data sets, and comment on the relationship between the two means.




15. If there are 8 numbers with a mean of 10 and

12 other numbers with a mean of 16, what is

the mean of all 20 numbers?

16. If the mean of 50 numbers is 18 and the mean

of the first 30 numbers is 16, what is the

mean of the last 20 numbers?

17. The mean of three numbers is 12 23 . If the

second number is one less than 3 times the

first, and the third is one more than twice the

second, what are the 3 numbers?

18. The mean of 50 numbers is 38. If two of the

numbers, namely 45 and 55, are removed,

what is the mean of the remaining numbers?


Foundations of Math 11 Section 5.2 – Standard Deviation ♦ 219


5.2 Standard Deviation

Consider the following question: 30 students are in the same Math and English classes; the mean for a test in Math and a test in English is 50%. If Mari scored 60% in Math and 70% in English, in which class did she do better compared to other students in the two classes? The answer to this question is that you cannot tell because you don’t know the spread or dispersion of test scores about the mean in the two classes. The value that measures the spread or dispersion of data about the mean is called standard deviation. The standard deviation is based on the deviations from the mean. We square the difference between each value and the mean. This squaring eliminates negative numbers. We then total the squared deviations, and divide the total by the number of values. The square root of this value is the standard deviation

Standard Deviation Definition σ = Greek symbol (sigma) for standard deviation of population.

σ = sum of the squares of the differences from the mean

number of values

=

(x1 − μ)2 + (x2 − μ)2 +� + (xn − μ)2

n (basic formula)

= 1

n(xi − μ)2

i=1

n

∑ (summation notation)

= 1

nxi

2 − μ2

i=1

n

∑ (alternate form of summation notation)




Example 1 Calculate the standard deviation from the following sets of values:

a) 5, 6, 7, 8, 9 b) 3, 5, 7, 9, 11

Solution:▼ Method 1 (by formula)

1. a) First find the mean: μ = 5+ 6 + 7 + 8+ 9

5 = 7

σ = 15 [(5− 7)2 + (6− 7)2 + (7 − 7)2 + (8− 7)2 + (9− 7)2 ]

=15 (4 +1+ 0 +1+ 4)

= 2

or

σ = 15 (52 + 62 + 72 + 82 + 92 )− 72

=15 (25+ 36 + 49 + 64 + 81)− 49

= 51− 49

= 2

1. b) First find the mean: μ = 3+ 5+ 7 + 9 +11

5 = 7

σ = 15 [(3− 7)2 + (5− 7)2 + (7 − 7)2 + (9− 7)2 + (11− 7)2 ]

=15 (16+ 4 + 0 + 4 +16)

= 2 2

or

σ = 15 (32 + 52 + 72 + 92 +112 )− 72

=15 (9+ 25+ 49 + 81+121)− 49

= 57 − 49

= 2 2

Notice that (3, 5, 7, 9, 11) are two units apart compared to one unit apart for (5, 6, 7, 8, 9). So the standard deviation is twice as large.





2. a) STAT (EDIT) ENTER (enter values under L1)

STAT (move cursor to CALC, then l : 1 – Var Stats) ENTER ENTER

2. b) STAT (EDIT) ENTER (enter values under L1)

STAT (move cursor to CALC, then 1 – Var Stats) ENTER ENTER

Note: A small standard deviation indicates the measures are clustered very close to the mean and a high standard deviation indicates the measures are widely scattered from the mean.

Standard Deviation

Standard Deviation




Example 2 Calculate the standard deviation for the following sets of data:

a) Daily Commute Time (minutes)

Number of Employees

0 to less than 10 4

10 to less than 20 9




25

b) Number of Orders

Number of Days

10–12 4

13–15 12

16–18 20

19–21 14

50


1. a) First find the mean:

μ = 5(4)+15(9) + 25(6) + 35(4) + 45(2)25

= 53525

= 21.4

σ = 4(5− 21.4)2 + 9(15− 21.4)2 + 6(25− 21.4)2 + 4(35− 21.4)2 + 2(45− 21.4)2

25= 11.6207

or

σ = 125

4 × 52 + 9 ×152 + 6 × 252 + 4 × 352 + 2 × 452⎡⎣ ⎤⎦ − 21.42

= 11.6207

1. b) First find the mean:

μ = 11(4)+14(12) +17(20)+ 20(14)50

= 83250

= 16.64

σ = 150 [4×112 +12 ×142 + 20 ×172 +14 × 202 ] –16.642

= 2.7259

or

σ = 4(11−16.64)2 +12(14 −16.64)2 + 20(17 −16.64)2 +14(20−16.64)2

50= 2.7259




Method 2

2. a) STAT (EDIT) ENTER (enter values under L1, L2)

STAT (move cursor to CALC, then 1 – Var Stats L1, L2) ENTER ENTER

Standard

Deviation

Mean

2. b) STAT (H:ClrList) ENTER (ClrList L1, L2) ENTER (Done) STAT (EDIT)

ENTER

STAT (move cursor to CALC, then 1 – Var Stats L1, L2) ENTER ENTER

Standard

Deviation

Mean




5.2 Exercise Set

1. The value of standard deviation is

a) never negative b) always positive c) never zero

2. Find the standard deviation of

2, 3, 5, 6 9

3. Find the standard deviation.

a) Score Frequency

1 1

2 3

3 5

4 4

5 2

b) Score Frequency

0 ≤ x < 10 1

10 ≤ x < 20 4

20 ≤ x < 30 3

30 ≤ x < 40 2

4. Find the standard deviation of the following

graph:

0

1

1

2

2

3

3

4

4

5

5

5. What do each of the following signify?

a) small standard deviation

b) large standard deviation

c) standard deviation of zero




6. All the required calculations to find the mean and standard deviation appear below:

Number of Orders

f (frequency)

m (midpoint) mf m – x (m – x)2 f (m – x )2

10–12 7 11 77 –5.28 27.8784 195.1488

13–15 12 14 168 –2.28 5.1984 62.3808

16–18 17 17 289 0.72 0.5184 8.8128

19–21 14 20 280 3.72 13.8384 193.7376

50 814 460.08

a) What is the mean? b) What is the standard deviation?




7. Fill in the table, and then find the standard deviation of 25 TVs sold during the past 25 weeks.

Number of Orders

f (frequency)

m (midpoint) mf m – x (m – x)2 f (m – x )2

0–5 2

6–10 6

11–15 10

16–20 4

21–25 3

8. Without doing any calculations, what is the

relationship between the standard deviation

of 1, 2, 3, 4, 5 and the standard deviation of

a) 3, 5, 7, 9, 11

b) 501, 502, 503, 504, 505

c) 21, 24, 27, 30, 33

9. Each of the following lists contains just one

value which is different from all the others.

Without calculating, what relationship is the

standard deviation between the values?

Set I: 6 6 6 10

Set II: 6 10 10 10




10. Each of the following lists contain an even number

of values of only two different values. Without

calculating, relate the standard deviation to the

difference between the values.

Set I: 16 16 20 20

Set II: 100 100 100 200 200 200

Set III: 60 60 60 60 80 80 80 80

11. The following scores are the results of

the midterm test given to five classes of

100 Calculus students at the University of

Saskatchewan. A score of 40 is needed to

pass, and a score of 85 is needed to get an A.

Group 1 2 3 4 5

Mean 50 50 60 70 70

Standard Deviation 6 14 10 18 2

Which of the five groups most likely

a) shows the highest student score?

b) shows the lowest student score?

c) shows the most uniform scores?

d) has the biggest range of scores?

e) has the most failures?

f) has the most A’s?

g) represents the overall average of all 5 classes?




μ

5.3 The Normal Distribution

If you gathered data from experiments such as the height, weight and test scores of individuals, it would follow a set pattern. This pattern, called a normal distribution, is extremely important because you only need the mean and standard deviation to complete the entire distribution. Important characteristics of the Normal Curve

• It is bell shaped and symmetric about the mean. • The enclosed area always equals one. • The probability that an event happens equals the area under the normal curve. • The standard deviation is related to the area under the curve. • The curve will never touch the x-axis but extends to infinity in both directions. • The mean, median and mode are always the same.

Two normal curves with the same standard deviation, but different means.

Two normal curves with the same mean, but different standard deviations.

1x 2x

x

Small standard deviation

Large standard deviation


Foundations of Math 11 Section 5.3 – The Normal Distribution ♦ 229


Z-score of Standard Normal Curve

There are many different possible normal curves with different values of μ and σ. By transforming each score into a z-score, which is a measure of how far a value is from the mean, the z-scores will fit the standard normal curve with μ = 0 and σ = 1 .

Definition

Z = difference between x and μ

standard deviation= x − μ

σ

Z: number of standard deviations that x is away from the mean μ

μ: mean x : a particular score σ : standard deviation

Note that there are just as many negative z-scores (when x < μ) as there are positive z-scores.

Standard Normal Curve

The value under the curve indicates that approximate proportion of area in each section. Note that over 68% of the data are within 1 standard deviation of the mean, and over 95% of the data are within 2 standard deviations. Relating Normal Curve and Standard Normal Curve

P(a ≤ X ≤ b) = P a − μ

σ< Z < b− μ

σ⎛⎝⎜

⎞⎠⎟

σμ−= aZ a

σμ−= bZb

μ

–3 σ –2 σ –1 σ 3 σ 2 σ 1 σ μ = 0




Example 1 If IQ scores are normally distributed with a mean of 100 and standard deviation of 15, determine

a) the z-score for 120 b) the probability that a randomly selected person has an IQ less than 120

Solution:▼ a) Z = x − μ

σ= 120 −100

15= 4

3 b)

P( X ≤120) = P Z < 120 −100

15⎛⎝⎜

⎞⎠⎟= P(Z < 4

3)

Method 1 (by table at back of chapter)

Find z-table in text, the table value always indicates area or probability to the LEFT of the z-score, therefore, P(Z < 1.33) = 0.9082. This score indicates that 90.82% of a normal population have an IQ less than 120.

Method 2 (by TI-83 calculator)

i) 2nd DISTR (move cursor down to normalcdf) ENTER (enter values)

ENTER

Or

normalcdf(left extreme, Zb ) normalcdf(left extreme, b, μ, σ)

Note: –1E99 indicates a lower bound value of the standard normal curve.

ii) WINDOW (enter values)

2nd DISTR (move cursor to

DRAW) ENTER (enter values)

ShadeNorm (–1E99 , 4/3) ENTER

Note: try to memorize these

max/min values for Z curve ShadeNorm(left extreme, Zb )




iii) WINDOW (enter values)

2nd DISTR (move cursor to DRAW)

ENTER (enter values) ShadeNorm

(–1E99, 120, 100, 15) ENTER

ShadeNorm(left extreme, b, μ, σ)

Note: If doing by ShadeNorm method, be sure to clear graph when finished by entering 2nd DRAW (ClrDraw) ENTER (DONE)

Example 2 The grade point average at Penticton Secondary is 2.6, with a standard deviation of 0.5. If the top 10% of all students are eligible to attend U.B.C., what is the minimum G.P.A. needed to attend U.B.C.? (Top 10% means that 90% will be less than the minimum G.P.A.)

Solution:▼

P( X ≤ 2.6) = P Z < x − 2.6

0.5⎛⎝⎜

⎞⎠⎟= 0.9 →

x − 2.6

0.5= Inverse Z0.9

Method 1 (by table)

Go to the Standard Normal Distribution Table and locate the closest value to 0.9 which is 0.8997. Once the closest number is found, read the value in the left column and top row to find the z-score value. The z-score for 0.8997 is 1.28.

Therefore, x − 2.6

0.5= 1.28 → x = 3.24. So, a G.P.A. of 3.24 is needed to go to U.B.C.


i) 2nd DISTR (move cursor to invNorm) ENTER (enter values) ENTER

Or

invNorm (area)

x − 2.6

0.5= 1.28 → x = 3.24

invNorm (area, μ , σ )

So, a G.P.A. of 3.24 is needed to go to U.B.C.

100 – 3 × 15100 – 3 × 15

– 0.1 ÷ 15

0.4 ÷ 15




5.3 Exercise Set

1. Find the area on the standard normal curve between – z and z if

a) z = 1

b) z = 2

c) z = 3

d) z = 4

e) z = 5

2. Decide what area under the standard normal curve is bigger or if the two areas are equal.

a) the area between z = – 1 and z = 1 or the area between z = 0 and z = 2

b) the area between z = 0.2 and z = 0.3 or the area between z = 1.2 and z = 1.3

c) the area between z = – 1 and z = – 0.5 or the area between z = 0.5 and z = 1

d) the area between z = 1 and z = 2 or the area between z = 2 and z = 4

e) the area to the right of z = – 2.5 or the area to the right of z = – 1.5




3. Find the area under the standard normal curve

a) between z = – 0.62 and z = 0.75

b) between z = – 2.35 and z = 1.42

c) between z = – 1.42 and z = – 2.38

d) to the right of z = 1.46

e) to the right of z = – 2.37

4. Find the area under a normal distribution curve with μ = 4 and σ = 10

a) area between x = – 3 and x = 9

b) area from x = 0 to x = 15

c) area to the right of x = – 6

d) area to the left of x = 2.31

e) area to the right of x = 12.42

5. If the probability of z < a = 0.8, determine a. 6. If the probability of x ≤ b = 0.6 and

μ = 4,σ = 10 , determine b.




7. The attendance for a week at the local

theatre is normally distributed, with a mean

of 4000 and a standard deviation of 500.

What percent of attendance figures fall

between 3600 and 4600 people?

8. Women pay on average $600 more for a car

than men. Assume a normal distribution of

charging with a mean of $600 and a standard

deviation of $50. Find the probability that a

woman pays at least $675 more than a man for

a car.

9. A manufacturer of cell phones indicated a

mean of 26 months before there is a need of

repairs, with a standard deviation of 6 months.

What length of time for the warranty should

the manufacturer set such that less than 10%

of all cell phones will need repairs during the

warranty period?

10. A provincial math exam has a mean of 68 and

a standard deviation of 13.2. If 30 000

students take the exam, and a score of 49 or

less fails, how many students fail the exam?




11. A normal random variable has a standard

deviation of 4. If the probability that X is less

than 15 is 0.75, what is the mean of X?

12. A normal random variable X has a mean of 80.

If the probability that X is less than 72 is 15%,

what is the standard deviation of X?

13. A test to a large group of students

approximates the normal curve. The mean is

70 with a standard deviation of 8. If 8% of

the students receive A’s, and 16% receive B’s,

what is the minimum mark needed to receive

a B?

14. At a high school, the average grade for

English is 64, with a standard deviation of 10.

If 20 students with grades between 73 and

85 receive B’s, how many students are taking

English at the high school?




5.4 Confidence Interval for Means

The concept of confidence intervals has to be one of the most important theories in all of statistics. A person often hears statements like:

• 1 of 5 Canadians is currently dieting • the life expectancy of smokers is five years less than non-smokers • the average teacher makes $43 000 a year

with results accurate to within two points, 19 times out of 20. Not every person in Canada is surveyed to obtain these results. One must then ask how many people were in the survey, and how accurate are the results. Sample measures are used to estimate population measures.

Confidence interval is defined as a specific interval estimate of the whole population by using information obtained from a sample, and the specific confidence level of the estimate.

It may be reasoned that the mean of the sample x should be centered about the mean of the whole population

μ. Because x is a single number, it is called a point estimate of the population mean μ. When the point estimate x is complete, the accuracy of the estimate is not really known. Statisticians have developed a method of taking a sample, and the standard normal distribution curve, to make an accurate estimate of the whole population.

STANDARD NORMAL DISTRIBUTION

left tail = 2

α right tail =

2

α

2

αZ− = μ – E

2

αZ = μ + E μ

E E

1 – α

x

For a given α, there is a probability of 1 – α that x will miss μ by less than the value of E = maximum error. The confidence interval is defined as:

x − E < μ < x + E for some population with maximum error

E = Zα

2

σn

Confidence Level Theorem μ = population mean

x = sample mean

Zα

2= standard deviation z–score separated into 2 tails

x − Zα

2

σn< μ < x + Zα

2

σn

where

n = sample size

⎧

⎨

⎪⎪⎪

⎩

⎪⎪⎪

σ = s = standard deviation of sample

Note: A sample size n > 30 must be used in this formula because σ, the standard deviation of the whole population and s, the standard deviation of a sample, are very close when n > 30.


Foundations of Math 11 Section 5.4 – Confidence Interval for Means ♦ 237


Example 1 A random sample of size 64 has a mean of 160 and a standard deviation of 15. Find a 95% confidence interval for the true mean of the population.

Note: Z0.025 = 1.96 from table

or invNorm (0.025)

or invNorm (0.975)

Solution:▼ Method 1 (by formula)

x − Zα

2

σn< μ < x + Zα

2

σn

160 − (1.96) 15

64

⎛⎝⎜

⎞⎠⎟< μ <160 + (1.96) 15

64

⎛⎝⎜

⎞⎠⎟

160 − 3.675 < μ <160 + 3.675

156.325 < μ <163.675

Method 2 (by TI-83 calculator)

STAT (move cursor to TESTS,

& down to ZInterval) ENTER

(enter values)

(move cursor to Calculate) ENTER

Which you read as 156.33 < μ < 163.67

Hence, you can say with 95% confidence that the true mean of the whole population is

between 156.33 and 163.67.

Or, as a reporter will say, the true mean of the population is 160, within 4, 19 times out of 20.




5.4 Exercise Set

1. The mean of the sampling distribution of x is always equal to

a) μ

b) Zα

2

c)

σn

2. The standard deviation of the sampling

distribution of the sample mean decreases

when

a) x increases

b) n increases

c) n decreases

3. When samples are selected from a normally

distributed population, the sampling

distribution of the sample mean has a normal

distribution

a) if n ≥ 30

b) if n ≥100

c) all the time

4. When samples are selected from a non-

normally distributed population, the sampling

distribution of the sample mean has a normal

distribution

a) if n ≥ 30

b) if n ≥100

c) all the time




5. Determine

Zα

2

for the following to two decimal places:

a) Zα

2

for 99% confidence interval

b) Zα

2


c) Zα

2


d) Zα

2


e) Zα

2





6. Determine the 95% confidence interval for

μ if σ = 6, x = 72 and n = 81 .


μ if σ = 6, x = 72 and n = 81 .


μ if σ = 6, x = 72 and n = 49 .


μ if σ = 6, x = 72 and n = 49 .




10. A study of 50 English teachers found the

average time spent marking a term paper

was 15.2 minutes with a standard deviation

of 2.8 minutes. Find a 94% confidence

interval of the mean time for all term papers.

11. Forest companies bid on a large tract of land

in the Prince George forest district. A random

sample of 150 trees yields a mean diameter

of 48 cm with a standard deviation of 5.6 cm.

Find a 90% confidence interval for the mean

diameter of all the trees.

12. A real-estate firm in Winnipeg takes a random

sample of 60 homes. This sample yields a

mean of 1800 square feet of living space with

a standard deviation of 280 square feet.

Construct a 99% confidence level for the mean

square footage of living space for all Winnipeg

homes.

13. Find the sample size necessary to estimate a

population mean to within 4 units if σ = 15 .

We want 90% confidence level in our results.




14. A study of 50 people living in Crescent Beach,

BC, showed the average age as 42 years with a

standard deviation of 12 years.

a) Find the 95% confidence interval of the

mean age for all the people living in

Crescent Beach.

b) If the 95% confidence interval of the study

stays the same, but we have 100 people

instead of 50, what happens to the

confidence interval? Why?

15. A random sample of 400 passengers that arrive

at Vancouver International Airport has a mean

processing time of 45 minutes with a standard

deviation of 12 minutes.

a) Construct a 98% confidence interval for the

mean arrival time for all passengers.

b) If the range you obtained in part a) is larger

than you want, how can you narrow it?

16. The confidence interval 78.82 < μ < 81.58 is

found by using a random sample for which

x = 80.2,σ = 5.3 and n = 40 . Determine the

degree of confidence.

17. A high school teacher wishes to estimate the

number of hours a student spends studying

each week. The standard deviation from a

previous study was 2.5 hours. How large a

sample must be selected if the teacher wants to

be 98% confident that the true mean differs

from the sample mean by 0.75 hours?


Foundations of Math 11 Section 5.5 – Chapter Review ♦ 243


5.5 Chapter Review

1. A population of scores is normally distributed with a mean of 62.4 and a standard deviation of 12.3.

If 40% of the scores are higher than a particular score x, calculate the value of x.

a) 59.3

b) 65.5

c) 69.6

d) 79.3

2. Determine the value of a in P(Z > a) = 0.7236 .

a) – 0.594

b) – 0.724

c) 0.594

d) 0.724

3. Determine the value of a in P(a < Z < 0) = 0.3925 .

a) – 2 .73

b) – 1.24

c) 1.24

d) 2.73




4. Let Z be a random variable with standard normal distribution. If P(a ≤ Z ≤ 2) = 0.1 , determine a.

a) 0.8773

b) 0.9773

c) 1.000

d) 1.161

5. Which frequency distribution shows a set of outcomes with the largest standard deviation?

a) b) c) d)

6. Determine the mean and standard deviation of the following graph.

a) μ = 2.87 , σ = 1.50

b) μ = 2.87 , σ = 1.58

c) μ = 3 , σ = 1.41

d) μ = 3 , σ = 1.50




Use the following information to answer questions 7 and 8.

Bed Day offers savings of $10, $25, $50 and $100 off on their already low price on bedroom suites. Select a coupon and save the amount on the coupon. There are a total of 100 coupons with 70, 20, 8 and 2 containing $10, $25, $50 and $100, respectively.

7. Determine the average saving.

a) $18.00

b) $25.00

c) $31.25

d) $46.25

8. Determine the standard deviation.

a) 16.46

b) 16.55

c) 34.16

d) 39.45




Use the following information to answer questions 9 and 10.

Let X be normally distributed with mean 6 and standard deviation 3.

9. Determine P(5 ≤ X ≤10) .

a) 0.539

b) 0.593

c) 0.625

d) 0.652

10. Determine b if P( X ≤ b) = 0.75 .

a) 3.98

b) 7.04

c) 8.02

d) 10.78

11. How does the mean of 9x, 10x, 11x, 12x compare to the mean of 3x, 4x, 5x, 6x?

a) 2 times as large

b) 2

13 times as large

c) 3 times as large

d) 6 units larger




12. How does the standard deviation of 4x, 6x, 8x, 10x compare to the standard deviation of x, 2x, 3x, 4x?

a) the same

b) 2 times as large

c) 4 times as large

d) 8 times as large

13. Determine the standard deviation of a fair die.

a) 1.708

b) 1.871

c) 3.5

d) 21

14. A teacher assigns grades in Mathematics according to the following procedure:

A if score exceeds μ + 1.3σ B if score between μ + 0.4σ and μ + 1.3σ C if score between μ – 0.5σ and μ + 0.4σ D if score between μ – 1.5σ and μ – 0.5σ E if score is below μ – 1.5σ If there are 30 students in the class, how many students receive C’s, assuming the scores are normally distributed?

a) 8

b) 10

c) 12

d) 14




15. A set of 500 scores are normally distributed. How many scores would you expect to find between 0.8 standard deviations and 2.5 standard deviations above the mean?

a) 100

b) 101

c) 102

d) 103

16. A four-point grading system gives 4 points for an A, 3 points for a B, 212 points for a C+, 2 points for a C,

and 1 point for a pass, with no points for a fail. If a student’s grades consist of 8 A’s, 15 B’s, 6 C+’s, 8 C’s,

2 passes and 1 fail, what is his grade point average?

a) 2.08

b) 2.50

c) 2.75

d) 2.82

17. The mean diameter of Okanagan apples is 12.5 cm with a standard deviation of 1.2 cm. If the diameter is normally distributed, what is the minimum diameter needed to only reject 10% of all apples?

a) 10.10 cm

b) 10.96 cm

c) 11.30 cm

d) 14.04 cm




18. From the list below, determine the minimum number of students with scores of 90 required to have the class

average over 75.

Score 60 70 80 90

Number of Students 7 3 8 ?

a) 5

b) 6

c) 7

d) 8

19. Find z if the standard normal curve area between – z and z is 0.4530.

a) ± 0.118

b) ± 0.398

c) ± 0.602

d) ± 0.752

20. The average height of adult males is 70 inches with a standard deviation of 2.4 inches. If the heights are

normally distributed, how high should a doorway be such that 95% of all men can pass through the doorway without hitting their heads.

a) 71 inches

b) 72 inches

c) 73 inches

d) 74 inches




21. If a random variable has the normal distribution with μ = 104.3 and σ = 5.7, find the probability that the value will be greater than 112.3.

a) 0.08

b) 0.09

c) 0.10

d) 0.11

22. In test A, a student received a mark of 43 with class mean 30 and standard deviation 10.

In test B, a student received a mark of 70 with class mean 60 and standard deviation 8. In test C, a student received a mark of 75 with class mean 60 and standard deviation 12.5. In test D, a student received a mark of 54 with class mean 50 and standard deviation 3. In which test did the student get the highest standard score?

a) A

b) B

c) C

d) D

23. The number of complaints about food in the school cafeteria per month is a random variable having the

normal distribution with μ = 8.3 and σ = 1.8, find the probability that in any month exactly 10 complaints will be reported.

a) 0.12

b) 0.13

c) 0.14

d) 0.15




24. In order to be a candidate for the R.C.M.P, recruits are given a stress test. The scores are normally

distributed with a mean of 62 and a standard deviation of 8.4. If just the top 25% of recruits are selected, determine the minimum score needed.

a) 66

b) 67

c) 68

d) 69

25. If the weight of female high school students closely follows the normal distribution with a mean of

125 pounds and a standard deviation of 5 pounds, what range of weights includes the middle 80% of the girls in high school?

a) 116.8 to 133.2

b) 118.6 to 131.4

c) 120.8 to 129.2

d) 122.3 to 127.1

26. In Math 11, the average grade is 64, and the standard deviation is 10. The teacher’s distribution has

8 students from 60.0 to 67.0 receiving C’s. Assuming normal distribution of grades, how many students are in the Math class?

a) 27

b) 29

c) 31

d) 33




27. Determine the standard deviation of the population a − 2d , a − d , a, a + d , a + 2d.

a) 2d

b) 3d

c) d

d) 2d

28. Most shoes last 2.4 years with a standard deviation of 1.2 years. A shoe company guarantees shoes for 1 year with free replacement. For every 500 shoes sold, assuming normal distribution, how many pairs of shoes must have free replacements?

a) 58

b) 59

c) 60

d) 61

29. One of Canada’s top universities only accepts the top 12% of all high school graduates on the basis of exam results. The exam results are normally distributed with a mean of 500 and a standard of 100. Determine the minimum score needed to enter this university.

a) 614

b) 616

c) 618

d) 620




30. A random variable has a normal distribution with σ = 3.0. If the probability is 0.945 that this random

variable will take on a value less than 85.2, what is the probability that it takes on a value greater than 78.4?

a) 0.6281

b) 0.6681

c) 0.7081

d) 0.7481




THE STANDARD NORMAL DISTRIBUTION TABLE

Area

0 z

Fz z( ) = P Z < z[ ]

z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09

–3.4 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0002 –3.3 0.0005 0.0005 0.0005 0.0004 0.0004 0.0004 0.0004 0.0004 0.0004 0.0003 –3.2 0.0007 0.0007 0.0006 0.0006 0.0006 0.0006 0.0006 0.0005 0.0005 0.0005 –3.1 0.0010 0.0009 0.0009 0.0009 0.0008 0.0008 0.0008 0.0008 0.0007 0.0007 –3.0 0.0013 0.0013 0.0013 0.0012 0.0012 0.0011 0.0011 0.0011 0.0010 0.0010

–2.9 0.0019 0.0018 0.0017 0.0017 0.0016 0.0016 0.0015 0.0015 0.0014 0.0014 –2.8 0.0026 0.0025 0.0024 0.0023 0.0023 0.0022 0.0021 0.0021 0.0020 0.0019 –2.7 0.0035 0.0034 0.0033 0.0032 0.0031 0.0030 0.0029 0.0028 0.0027 0.0026 –2.6 0.0047 0.0045 0.0044 0.0043 0.0041 0.0040 0.0039 0.0038 0.0037 0.0036 –2.5 0.0062 0.0060 0.0059 0.0057 0.0055 0.0054 0.0052 0.0051 0.0049 0.0048

–2.4 0.0082 0.0080 0.0078 0.0075 0.0073 0.0071 0.0069 0.0068 0.0066 0.0064 –2.3 0.0107 0.0104 0.0102 0.0099 0.0096 0.0094 0.0091 0.0089 0.0087 0.0084 –2.2 0.0139 0.0136 0.0132 0.0129 0.0125 0.0122 0.0119 0.0116 0.0113 0.0110 –2.1 0.0179 0.0174 0.0170 0.0166 0.0162 0.0158 0.0154 0.0150 0.0146 0.0143 –2.0 0.0228 0.0222 0.0217 0.0212 0.0207 0.0202 0.0197 0.0192 0.0188 0.0183

–1.9 0.0287 0.0281 0.0274 0.0268 0.0262 0.0256 0.0250 0.0244 0.0239 0.0233 –1.8 0.0359 0.0352 0.0344 0.0336 0.0329 0.0322 0.0314 0.0307 0.0301 0.0294 –1.7 0.0446 0.0436 0.0427 0.0418 0.0409 0.0401 0.0392 0.0384 0.0375 0.0367 –1.6 0.0548 0.0537 0.0526 0.0516 0.0505 0.0495 0.0485 0.0475 0.0465 0.0455 –1.5 0.0668 0.0655 0.0643 0.0630 0.0618 0.0606 0.0594 0.0582 0.0571 0.0559

–1.4 0.0808 0.0793 0.0778 0.0764 0.0749 0.0735 0.0722 0.0708 0.0694 0.0681 –1.3 0.0968 0.0951 0.0934 0.0918 0.0901 0.0885 0.0869 0.0853 0.0838 0.0823 –1.2 0.1151 0.1131 0.1112 0.1093 0.1075 0.1056 0.1038 0.1020 0.1003 0.0985 –1.1 0.1357 0.1335 0.1314 0.1292 0.1271 0.1251 0.1230 0.1210 0.1190 0.1170 –1.0 0.1587 0.1562 0.1539 0.1515 0.1492 0.1469 0.1446 0.1423 0.1401 0.1379

–0.9 0.1841 0.1814 0.1788 0.1762 0.1736 0.1711 0.1685 0.1660 0.1635 0.1611 –0.8 0.2119 0.2090 0.2061 0.2033 0.2005 0.1977 0.1949 0.1922 0.1894 0.1867 –0.7 0.2420 0.2389 0.2358 0.2327 0.2296 0.2266 0.2236 0.2206 0.2177 0.2148 –0.6 0.2743 0.2709 0.2676 0.2643 0.2611 0.2578 0.2546 0.2514 0.2483 0.2451 –0.5 0.3085 0.3050 0.3015 0.2981 0.2946 0.2912 0.2877 0.2843 0.2810 0.2776

–0.4 0.3446 0.3409 0.3372 0.3336 0.3300 0.3264 0.3228 0.3192 0.3156 0.3121 –0.3 0.3821 0.3783 0.3745 0.3707 0.3669 0.3632 0.3594 0.3557 0.3520 0.3483 –0.2 0.4207 0.4168 0.4129 0.4090 0.4052 0.4013 0.3974 0.3936 0.3897 0.3859 –0.1 0.4602 0.4562 0.4522 0.4483 0.4443 0.4404 0.4364 0.4325 0.4286 0.4247 –0.0 0.5000 0.4960 0.4920 0.4880 0.4840 0.4801 0.4761 0.4721 0.4681 0.4641




Fz z( ) = P Z < z[ ]

z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09

0.0 0.5000 0.5040 0.5080 0.5120 0.5160 0.5199 0.5239 0.5279 0.5319 0.5359 0.1 0.5398 0.5438 0.5478 0.5517 0.5557 0.5596 0.5636 0.5675 0.5714 0.5753 0.2 0.5793 0.5832 0.5871 0.5910 0.5948 0.5987 0.6026 0.6064 0.6103 0.6141 0.3 0.6179 0.6217 0.6255 0.6293 0.6331 0.6368 0.6406 0.6443 0.6480 0.6517 0.4 0.6554 0.6591 0.6628 0.6664 0.6700 0.6736 0.6772 0.6808 0.6844 0.6879

0.5 0.6915 0.6950 0.6985 0.7019 0.7054 0.7088 0.7123 0.7157 0.7190 0.7224 0.6 0.7257 0.7291 0.7324 0.7357 0.7389 0.7422 0.7454 0.7486 0.7517 0.7549 0.7 0.7580 0.7611 0.7642 0.7673 0.7704 0.7734 0.7764 0.7794 0.7823 0.7852 0.8 0.7881 0.7910 0.7939 0.7967 0.7995 0.8023 0.8051 0.8078 0.8106 0.8133 0.9 0.8159 0.8186 0.8212 0.8238 0.8264 0.8289 0.8315 0.8340 0.8365 0.8389

1.0 0.8413 0.8438 0.8461 0.8485 0.8508 0.8531 0.8554 0.8577 0.8599 0.8621 1.1 0.8643 0.8665 0.8686 0.8708 0.8729 0.8749 0.8770 0.8790 0.8810 0.8830 1.2 0.8849 0.8869 0.8888 0.8907 0.8925 0.8944 0.8962 0.8980 0.8997 0.9015 1.3 0.9032 0.9049 0.9066 0.9082 0.9099 0.9115 0.9131 0.9147 0.9162 0.9177 1.4 0.9192 0.9207 0.9222 0.9236 0.9251 0.9265 0.9278 0.9292 0.9306 0.9319

1.5 0.9332 0.9345 0.9357 0.9370 0.9382 0.9394 0.9406 0.9418 0.9429 0.9441 1.6 0.9452 0.9463 0.9474 0.9484 0.9495 0.9505 0.9515 0.9525 0.9535 0.9545 1.7 0.9554 0.9564 0.9573 0.9582 0.9591 0.9599 0.9608 0.9616 0.9625 0.9633 1.8 0.9641 0.9649 0.9656 0.9664 0.9671 0.9678 0.9686 0.9693 0.9699 0.9706 1.9 0.9713 0.9719 0.9726 0.9732 0.9738 0.9744 0.9750 0.9756 0.9761 0.9767

2.0 0.9772 0.9778 0.9783 0.9788 0.9793 0.9798 0.9803 0.9808 0.9812 0.9817 2.1 0.9821 0.9826 0.9830 0.9834 0.9838 0.9842 0.9846 0.9850 0.9854 0.9857 2.2 0.9861 0.9864 0.9868 0.9871 0.9875 0.9878 0.9881 0.9884 0.9887 0.9890 2.3 0.9893 0.9896 0.9898 0.9901 0.9904 0.9906 0.9909 0.9911 0.9913 0.9916 2.4 0.9918 0.9920 0.9922 0.9925 0.9927 0.9929 0.9931 0.9932 0.9934 0.9936

2.5 0.9938 0.9940 0.9941 0.9943 0.9945 0.9946 0.9948 0.9949 0.9951 0.9952 2.6 0.9953 0.9955 0.9956 0.9957 0.9959 0.9960 0.9961 0.9962 0.9963 0.9964 2.7 0.9965 0.9966 0.9967 0.9968 0.9969 0.9970 0.9971 0.9972 0.9973 0.9974 2.8 0.9974 0.9975 0.9976 0.9977 0.9977 0.9978 0.9979 0.9979 0.9980 0.9981 2.9 0.9981 0.9982 0.9982 0.9983 0.9984 0.9984 0.9985 0.9985 0.9986 0.9986

3.0 0.9987 0.9987 0.9987 0.9988 0.9988 0.9989 0.9989 0.9989 0.9990 0.9990 3.1 0.9990 0.9991 0.9991 0.9991 0.9992 0.9992 0.9992 0.9992 0.9993 0.9993 3.2 0.9993 0.9993 0.9994 0.9994 0.9994 0.9994 0.9994 0.9995 0.9995 0.9995 3.3 0.9995 0.9995 0.9995 0.9996 0.9996 0.9996 0.9996 0.9996 0.9996 0.9997 3.4 0.9997 0.9997 0.9997 0.9997 0.9997 0.9997 0.9997 0.9997 0.9997 0.9998


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