Suppose you take a sample of 200 lehman students. the average height is 65 inches (5’4"") with a standard deviation of 5 inches. using the standard normal distribution: a) what would be the height in inches when the probability is 4.95% (0.0495) and when the probability is 95.05% (0.9505)? b) what would be the height if the probability is 99.01% (0.9901)?

Step-by-step explanation:

Using the Standard Normal Distribution, the formula for normal distribution is expressed as

z = (x - u)/s

Where

x = height of students

u = mean score

s = standard deviation

From the information given,

n = 200

u = 65 inches

Standard deviation = 5 inches

a) when the probability is 4.95%. p = 0.0495

From the normal distribution table,

z = - 1.65

Therefore

- 1.65 = ( x- 65)/5

x - 65 = 5 × - 1.65 = - 8.25

x = -8.25 + 65 = 56.75 inches

The height is 56.75 inches

when the probability is 95.05%. p = 0.9505

From the normal distribution table,

z = 1.65

Therefore

1.65 = ( x- 65)/5

x - 65 = 5 × 1.65 = 8.25

x = 8.25 + 65 = 73.25 inches

The height is 73.25 inches

b) when the probability is 99.01%.

p = 0.9901

From the normal distribution table,

z = 2.33

Therefore

2.33 = ( x- 65)/5

x - 65 = 5 × 2.33 = 11.65

x = 11.65 + 65 = 76.65 inches

The height is 76.65 inches

it can be written as 4(p+q)