2-5 : Normal Distribution

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Standard Deviation

Given a Data Set 12, 8, 7, 14, 4

The standard deviation is a measure of the mean spread of

the data from the mean.

Mean = (12 + 8 + 7 + 14 + 4) ÷ 5 = 9x

Calculate the mean

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xx 2

How far is each data value from the mean?

Square to remove the negatives

Average = Sum divided by how many values

Square root to ‘undo’ the squared

(25 + 4 + 25 + 1 + 9) ÷ 5 = 12.8

Square root 12.8 = 3.58

Std Dev = 3.58

nCalculator function

NORMAL DISTRIBUTIONS : Vocabulary

Normal curves are used in many situations to estimate probabilities. The pattern the data form is a bell-shaped curve called a normal curve.

NORMAL DISTRIBUTION ̶ shows data that vary randomly from the mean in the pattern of a bell-shaped curve.

NORMAL DISTRIBUTIONS : Vocabulary

1. Maximum value of the curve occurs at the mean, usually symbolized as , (Greek letter mu). 𝜇2. The standard deviation of (sigma). 𝜎3. Nearly all of the area under the curve (99.7%) is within three standard deviations from the mean,

±3 𝜇 𝜎4. The total area under a normal curve is always 1.

5. The area under the curve between two x-values corresponds to the probability that x is between those values.

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Properties of Normal Distributions

If x is a continuous random variable having a normal distribution with mean μ and standard deviation σ.

μ 3σ μ + σμ 2σ μ σ μ μ + 2σ μ + 3σ

Total area = 1

x

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Z-Scores

EX #7: Use the data from EX #4. Plot the data values and the mean on the line plot.

A. Do all of the values fall between two standard deviations of mean?

No : 32 MWH is more than 2 standard deviation below the mean.

B. In August, the mean daily energy requirements is 39.4 MWh, with a standard deviation of 3.2 MWh. The power company makes plans for any demand within three standard deviations of the mean. Will they be prepared for a demand of 51 MWh? Explain.

39.4 + 3(3.2) = 49 ; No, 51 is an outlier.

C. Find the z-score. = 2.7

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Estimating Probabilities Using A Normal Curve

EX #1: In a survey of American men, the heights are distributed normally about the mean, as shown in the graph above.

A. About what percent of men aged 26 to 35 are 68 – 70 inches tall? 11 + 13.2 +16 40.2%

B. About what percent of the men in the survey are less than 70 inches tall? 1+3+5.8+6+11+13.2 40%C. Suppose this survey was the data on 1500 men. About how many men would you expect to be 68 – 70 inches tall? (0.402)(1500)

70 + 2.5 = 67.5 70 - 2.5 = 72.5 11 + 13.2 +16 +14.2+12 66.4%

D. The mean of the data is 70, and the standard deviation is 2.5. About what percent of the men are within one standard deviation of the mean in height?

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Sketching a Normal CurveEX #2: The table below shows the lengths, in feet, of 40 great white sharks from Scuba News Magazine.

A. Enter these data items into your calculator and make a histogram

B. Calculate one-variable statistics.

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Sketching a Normal CurveC. Sketch a normal curve, showing the frequency of the shark lengths. Label the x-axis values at one, two, and three standard deviations from the mean.

μ 3σ8.2

μ + σ18.2

μ 2σ10.7

μ σ13.2

μ15.7

μ + 2σ20.7

μ + 3σ23.2

Total area = 1

x51

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The Normal DistributionKey Concepts

Total Area = 1

x

Distributions with different spreads have different

STANDARD DEVIATIONS

Area under the graph is the relative frequency = the probability

The MEAN is in the middle.The distribution is symmetrical.

A lower mean

x

A higher mean

x

A smaller Std Dev.

A larger Std Dev.

1 Std Dev either side of mean = 68%

2 Std Dev either side of mean = 95%

3 Std Dev either side of mean = 99%

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The 68 – 95 – 99.7 Rule When a data set is normally distributed, about 68% of the data fall within one standard deviation of the mean. About 95% of the data fall within two standard deviations of the mean.

In order to find the values that are two standard deviations away from the mean, we would find the values that have a z-score of ±2. Sometimes called The Empirical Rule:

The 68 – 95– 99.7 Rule In a Normal distribution with mean and standard deviation : 𝜇 𝜎• Approximately 68% of the observations fall within of the mean. 𝜎• Approximately 95% of the observations fall within 2 of the mean. 𝜎• Approximately 99.7% of the observations fall within 3 of the 𝜎

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Using Standard Normal Values If a random variable x is normally distributed with mean 𝜇 and standard deviation 𝜎𝜎, then the standard normal value of x is

The table shows the approximate area under the curve for all values less than z for selected values of z.

EX #3: Scores on a test are normally distributed with a mean of 76 and a standard deviation of 8. A. Estimate the probability that a randomly selected student scored less than 88%. = 1.5 Find area under the curve for all values less than 1.5 or about 0.93

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Using Standard Normal Values

B. Estimate the probability that a randomly selected student scored between 72 and 80.

= = = -0.5 = 0.5

Area Area

Since these area overlap subtract 0.69-0.31 = 0.38

μ 3σ μ + σμ 2σ μ σ μ μ + 2σ μ + 3σ

Total area = 1

x

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SkewnessSkewness

NegativelySkewed

PositivelySkewed

Symmetric(Not Skewed)

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SkewnessSkewness

NegativelySkewed

Mode

Median

Mean

Symmetric(Not Skewed)

MeanMedianMode

PositivelySkewed

Mode

Median

Mean

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Is It Normal or Skewed?

While many data sets can be modeled using a normal curve, not all data is normally distributed. Sometimes the “tail” is longer on one side than the other, resulting in a skewed distribution.

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Is It Normal or Skewed?EX #4: The heights of 24 children entering a theme park are shown below. The mean of all the children at the park is 49 inches and the standard deviation is 6 inches.

A. Does the data appear to be normally distributed? Explain your reasoning.

No, this data is skewed positively or left. There are more values above 49 inches than below.

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B. Use your calculator to draw a histogram and calculate one-variable statistics.

Is It Normal or Skewed?

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Real-World Connection EX #5: In a college algebra course with 286 students in a lecture hall, the final exam scores have a mean of 67.5 and a standard deviation of 7.4. The grades on the exams are all whole numbers, and the grade pattern follows a normal curve.

A. Sketch a normal curve, label the x-axis values at one, two, and three standard deviations from the mean.

μ 3σ = 45.3

μ + σ = 74.9μ 2σ = 52.7

μ= 67.5 μ + 2σ = 89.7 μ + 3σ

Total area = 1

xμ σ = 60.1

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397

397

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Real-World Connection B. Find the number of students who receive grades from one to two standard deviations above the mean.

0.135(285) = 38.61

C. How many students earned a grade below 60%?

13.5% +2.5% = 16%0.16(286) = 4.76About 46 students earned 60 or below.

D. Suppose the class only had 170 students. About how many students would earn a grade of 82 or more? 13.5% +2.5% = 16%0.16(170) = 27.2