Question

According to the U.S. National Center for Health Statistics, 25.2 percent of males and 23.6 percent of females never eat breakfast. Suppose that random samples of 200 men and 200 women are chosen. Approximate the probability that

(a) at least 110 of these 400 people never eat breakfast;

(b) the number of the women who never eat breakfast is at least as large as the number of the men who never eat breakfast.

Short answer

(a) Probability: $P\{110\le M+N\}=0.0749$

(b) Probability :

$P\{M\le W\}=0.3557$

Step-by-step solution

Introduction

The National Center for Health Statistics' (NCHS) objective is to produce statistical data that will influence activities and policies aimed at improving Americans' health. NCHS, as the nation's primary health statistics agency, sets the standard for reliable, timely, and relevant data.

Given 

$25.2\%$of males never eat breakfast.

$23.6\%$of females never eat breakfast.

200 men and 200 women are chosen at random.

Explanation (a)

Let

M denote the number of men that never eat breakfast

W denote the number women that never eat breakfast

Note that

M is a binomial random variable

With

$p=0.252$and $n=200$

$W$is a binomial random variable

With

$p=0.236andn=200$

Then compute

$E\left[M\right]=200(0.252)=50.4$

And

$E\left[W\right]=200(0.236)=47.2$

Also,

$Var\left[M\right]=200(0.252)(1-0.252)\approx 37.7$

And

$Var\left[W\right]=200(0.236)(1-0.236)\approx 36.1$

Thus,

We need to approximate

M by a normal random variable

With

$\mu =50.4\text{and}{\sigma }^2\approx 37.7$

W by a normal random variable

With

$\mu =47.2\text{and}{\sigma }^2\approx 36.1$

Now,

We need to compute $P\{110\le M+W\}$.

Approximate $M+W$by a nomal distribution

With

$\mu =50.4+47.2=97.6$

And

${\sigma }^2\approx 37.7+36.1=73.8$

Thus,

$P=\{110\le M+W\}=0.0749$

Explanation (b)

Let

M denote the number of men that never eat breakfast

W denote the number women that never eat breakfast

Note that

M is a binomial random variable

With

$p=0.252\text{and}n=200$

$W$is a binomial random variable

With

$p=0.236\text{and}n=200$

Then compute

$E\left[M\right]=200(0.252)=50.4$

And

$E\left[W\right]=200(0.236)=47.2$

Also,

$Var\left[M\right]=200(0.252)(1-0.252)\approx 37.7$

And

$Var\left[W\right]=200(0.236)(1-0.236)\approx 36.1$

Thus,

We need to approximate

M by a normal random variable

With

$\mu =50.4\text{and}{\sigma }^2\approx 37.7$

W by a normal random variable

With

$\mu =47.2\text{and}{\sigma }^2\approx 36.1$

Now,

We need to compute $P\{M\le W\}=P\{M-W\le 0\}$.

Approximate $M-W$by a normal distribution

With

$\mu =50.4-47.2=3.2$

And

${\sigma }^2\approx 37.7+36.1=73.8$

Thus,

$P\{M\le W\}=0.3557$