In a particular area 45% of men and 10% of women are overweight and there are sixmen and four women working in an office there. Find the probability that there are 2overweight people in the office. Express you answer in 4 decimal places.

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Description: In a particular area 45% of men and 10% of women are overweight and there are sixmen and four women working in an office there. Find the probability that there are 2overweight people in the office. Express you answer in 4 decimal places.

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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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In a particular area 45% of men and 10% of women are overweight and there are six
men and four women working in an office there. Find the probability that there are 2
overweight people in the office. Express you answer in 4 decimal places.

Expert Solution

Step 1

Given Information:

45% of men and 10% of women are overweight.

There are six men and four women in an office.

To find the probability that there are 2 overweight people in the office.

Let $p = 0.45$ be the probability of success i.e., probability that men are overweight.

$n = 6$

Then, X be a Binomial random variable which denotes the number of overweight men.

Binomial formula:

$P (X = x) = {}^{n}C_{x} p^{x} {(1 - p)}^{n - x}$

Probability that there are 2 overweight people in the office can be obtained as follows:

$P (2 o v e r w e i g h t p e o p l e) = P (e x a c t l y 2 o v e r w e i g h t m e n a n d n o o v e r w e i g h t w o m e n) + P (e x a c t l y 2 o v e r w e i g h t w o m e n a n d n o o v e r w e i g h t m e n) + P (e x a c t l y 1 o v e r w e i g h t m a n a n d 1 o v e r w e i g h t w o m a n)$

Step 2

Probability that exactly 2 men are overweight is calculated as follows:

$\begin{array}{rcl} P (X = 2) & = & {}^{6}C_{2} {(0.45)}^{2} {(1 - 0.45)}^{6 - 2} \\ = & 15 \times 0.2025 \times 0.09150625 \\ = & 0.2779502344 \\ \approx & 0.27795 \end{array}$

Probability that exactly 1 man is overweight is calculated as follows:

$\begin{array}{rcl} P (X = 1) & = & {}^{6}C_{1} {(0.45)}^{1} {(1 - 0.45)}^{6 - 1} \\ = & 6 \times 0.45 \times 0.0503284375 \\ = & 0.1358867813 \\ \approx & 0.13589 \end{array}$

Probability that no men are overweight is calculated as follows:

$\begin{array}{rcl} P (X = 0) & = & {}^{6}C_{0} {(0.45)}^{0} {(1 - 0.45)}^{6 - 0} \\ = & 1 \times 1 \times 0.02768064063 \\ = & 0.02768064063 \\ \approx & 0.02768 \end{array}$

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