Question

Setup: The heights of all students in Richland County have been found to follow a Normal Distribution with a mean of 66 inches and a standard deviation of 2.5 inches.

1.) Find the percent of the RCSD student population tat is more than 68 inches tall.

2.) If I were to select 3 students in a row, what is the probability that all 3 of them would be more than 68 inches tall?

3.) Becky calls home and proudly announces that only 35% of all RCSD students are taller than she is. How tall is Becky?

4.) Out of the 160,000 students in Richland County, how many would you expect to have heights within two standard deviations of the mean?

Answer 1

Given = 66, = 2.5

To find the probability, we need to find the z scores.

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(1) n = 1 , For P (X > 68) = 1 - P (X < 68), as the normal tables give us the left tailed probability only.

For P( X < 68)

Z = (68 – 66)/2.5 = 0.8

The probability for P(X < 68) from the normal distribution tables is = 0.7881

The required probability = 1 – 0.7881 = 0.2119

The percentage is therefore 0.2119 * 100 = 21.19%

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(2) n = 3 , For P (X > 68) = 1 - P (X < 68), as the normal tables give us the left tailed probability only.

For P( X < 68)

Z = (68 – 66)/[2.5/sqrt(3)] = 1.39

The probability for P(X < 68) from the normal distribution tables is = 0.9177

The required probability = 1 – 0.9177 = 0.0823

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(3) There are 65% students who are below beck

Therefore P(X < x) = 0.65

The z score at p = 0.65 is 0.3853

Therefore (X - 66) / 2.5 = 0.3853

Solving for X, we get X = (0.3853 * 2.5) + 66 = 66.96 inches

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(4) 2 standard deviations = 2.5 * 2 = 5

Therefore P(66 - 5 < X < 66 + 5) = P(61 < X < 71) = P(X < 71) - P(X < 61)

For P( X < 71)

Z = (71 – 66)/2.5 = 2

The probability for P(X < 71) from the normal distribution tables is = 0.9772

For P( X < 61)

Z = (61 – 66)/2.5 = -2

The probability for P(X < 61) from the normal distribution tables is = 0.0228

The proportion of students between 2 standard deviations 0.9772 – 0.0228 = 0.9544

The number of students = 160000 * 0.9544 = 152704

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