percentage points.pdf

AQA Statistics 1 The normal distribution

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Section 2: Percentage points Notes and Examples These notes contain subsections on

Working backwards

Non-standardised variables

Further examples

Working backwards If you need to find the value of z associated with a particular probability, then you need to work backwards. To do this, you need to be confident using the tables giving percentage points of the Normal distribution. Since the inverse Normal tables start with a probability of 0.5, you must always work with a probability of 0.5 or above. If you are given a probability of less than 0.5, you need to use symmetry:

P(Z < a) = p P(Z < -a) = 1 p Or, using the alternative notation:

(a) = p (-a) = 1 p Note that this is only true for a standard Normal distribution, since the mean is zero. Example 1 Version 1

Find a and b where:

(i) P(Z < a) = 0.85

(ii) P(Z < b) = 0.15

Solution

Using the table for the percentage points of the Normal distribution:

(i) Looking up p = 0.85 gives z = 1.036

So a = 1.036

(ii) By symmetry P(Z < b) = 0.15 P(Z < -b) = 0.85 From part (i) you can see that -b = 1.036.

So b = -1.036

Example 1 Version 2

Find a and b where:

(i) (a) = 0.85

(ii) (b) = 0.15

AQA S1 Normal distribution 2 Notes and Examples

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Solution

Using the table for the percentage points of the Normal distribution:

(i) 1(0.85) 1.036 a

(ii) By symmetry P(Z < b) = (b) = 0.15 (-b) = 0.85

1(0.85) 1.036 b

So b = -1.036

Non-standardised variables Once you are confident using standardised normal variables you can apply the same techniques to other normal distributions by standardising the

variables using x

z

.

To convert from standardised scores to the value in its original context, you can use x z .

Example 2 Version 1

Assume a test is normally distributed with a mean of 50 and a standard deviation of 6.

Let X be the distribution of test scores.

Find a and b where:

(i) P(X > a) = 0.39

(ii) P(X < c) = 0.15

Solution

X N (50, 62)

(i) P(X > a) = 0.39 P(X < a) = 0.61 From tables: P(Z < 0.2793) = 0.61

So z = 0.2793

Using x z : a = 50 + 6 0.2793 = 51.68

(ii) By symmetry P(Z < z) = 0.15 P(Z < -z) = 0.85 From the tables P(Z < 1.036) = 0.85

So z = -1.036

Using x z : b = 50 + 6 1.036 = 43.78

Solution 2

X N (50, 62)

(i) P(X > a) = 0.39 P(X < a) = 0.61

From tables 1(0.612) 0.2793

Using x z : a = 50 + 6 0.2793 = 51.68


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(ii) P(X < b) = 0.15

By symmetry P(Z < z) = 0.15 P(Z < -z) = 0.85 1(0.85) 1.036

1.036

z

z

Using x z : b = 50 + 6 1.036 = 43.78

Further examples A similar approach is undertaken if the value of or is unknown and is

illustrated in the example below where has to be found.

Example 3

Assume a test is normally distributed with a mean of 50 and a standard deviation of

. Let X be the distribution of test scores. Find if the probability of getting a score above 66 is 0.11.

Solution 1

X N (50, 2)

P(X > 66) = 0.11 P(X < 66) = 0.89 From tables, P(Z < 1.227) = 0.89

So z = 1.227

Now when x = 66, 66 50 16

z

161.227

= 16

13.041.227

Solution 2

X N (50, 2)

P(X > 66) = 0.11 P(X < 66) = 0.89 1( ) 0.89 (0.89) 1.227 z z

Now when x = 66, 66 50 16

z

161.227

1613.04

1.216

The final example looks at determining the limits where a percentage of the distribution is expected to lie.


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Example 4

Consider a Normal variable with mean 70 and variance 25.

Find the limits within which the central 95% of the distribution lies.

Solution

X N(70, 25)

Let the limits within which the central 95% of the

distribution lies be a and b, where a < b. If 95% of the

values lie within a and b, then 2.5% of the values lie

either side of a and b.

P(X < b) = 0.975

From tables P(Z < 1.96) = 0.975 so z = 1.96

Using x z : b = 70 + 5 1.96 = 79.8

EITHER: Now a and b must be symmetrical about the mean, which is 70. a = 70 5 1.96 = 60.2

OR: P(X < a) = 0.025

From tables P(Z < -1.96) = 0.025 so z = -1.96

Using x z : a = 70 5 1.96 = 60.2

Solution 2

X N (70, 25)

Let the limits within which the central 95% of the distribution lies be a and b, where

a < b. If 95% of the values lie within a and b, then 2.5% of the values lie either side of

a and b.

P(X < b) = 0.975

From tables 1(0.975) 1.96

Using x z : b = 70 + 51.96 = 79.8

EITHER: Now a and b must be symmetrical about the mean, which is 70. a = 70 51.96 = 60.2

OR: P(X < a) = 0.025

From tables 1(0.025) 1.96

Using x z : a = 70 - 51.96 = 60.2

70 b a

95%

2.5% 2.5%

percentage points.pdf

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