Hi, I need help with this please. 8. over 6 feet. (72 inches)9. under 65 inches.10. between 62 inches and 65 inches.

Name: Hi, I need help with this please. 8. over 6 feet. (72 inches)9. under
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Description: Hi, I need help with this please. 8. over 6 feet. (72 inches)9. under 65 inches.10. between 62 inches and 65 inches.

Answered: 8. over 6 feet. (72 inches) 9. under 65…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Hi, I need help with this please.

8. over 6 feet. (72 inches)
9. under 65 inches.
10. between 62 inches and 65 inches.

Expert Solution

Step 1

The random variable X is defined as the height of UNLV female student which follows normal distribution with mean 63.7 inches and a standard deviation of 2.3 inches.

The probability that a UNLV female student with over 6 feet(72 inches) is,

$\begin{array}{rcl} P (X > 72) & = & 1 - P (X < 72) \\ = & 1 - P (\frac{X - 63.7}{2.3} < \frac{72 - 63.7}{2.3}) \\ = & 1 - P (z < 3.61) \end{array}$

The probability of z less than 3.61 can be obtained using the excel formula “=NORM.S.DIST(3.61,TRUE)”.

The probability value is 0.9998.

The required probability is,

$\begin{array}{rcl} P (X > 72) & = & 1 - P (z < 3.61) \\ = & 1 - 0.9998 \\ = & 0.0002 \end{array}$

Thus, the probability that a UNLV female student with over 6 feet(72 inches) is 0.0002.

Step 2

The probability that a UNLV female student with under 65 inches is,

$\begin{array}{rcl} P (X < 65) & = & P (\frac{X - 63.7}{2.3} < \frac{65 - 63.7}{2.3}) \\ = & P (z < 0.565) \end{array}$

The probability of z less than 0.565 can be obtained using the excel formula “=NORM.S.DIST(0.565,TRUE)”.

The probability value is 0.7140.

The required probability is,

$\begin{array}{rcl} P (X < 65) & = & P (z < 0.565) \\ = & 0.7140 \end{array}$

Thus, the probability that a UNLV female student with under 65 inches is 0.7140.

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