Question

The following table gives the percentage of individuals, categorized by gender, that follow negative practices.

Sleep 6 Hours

or Less per Night Smoker

Men 22.7 28.4 Women 21.4 22.8

Rarely Eats Breakfast 45.4 42.0

Is 20 Percent or More Overweight 29.6

25.6

a) Suppose a random sample of 300 men is chosen, approximate the probability that at least 150 of them rarely eat breakfast.

b) Suppose a random sample of 300 women is chosen, approximate the probability that fewer than 50 of them sleep 6 hours or less nightly.

Answer 1

Answer:

(A)

Given,

To determine the probability that atleast 150 of them rarely eat breakfast

n =300

percentage of rarely eat breakfast for men = 45.4 %

p = 0.454

Now,

np = 300* 0.454

= 136>5

p+q = 1

q=1-p

nq = 300* 0.546

= 164> 5

mean = np

= 300*0.454

= 136.2

standard deviation

= sqrt(300*0.454*0.546)

Standard deviation = 8.62

Now probability

i.e., P(X=> 150) = P(X> 149.5)

P(X> 149.5) = P{Z=> (149.5 -136.2)/8.62)

= P(Z >= 1.54)

= 1-P( <=1.54)

= 1- 0.9382 [since from the normal distribution table]

P(X> 149.5) = 0.0618

(B)

Given,

To determine the probability that fewer than 50 of them sleep 6 hours or less nightly

percentage of the women sleep 6 hours or less than = 0.214

Consider,

np = 300*0.214

= 64 >5

nq = 300* 0.690

= 207 >5

we can use normal approximation

mean = np

= 300* 0.214

= 64.2

standard deviation

standard deviation = sqrt(300*0.214*0.690)

standard deviation = 6.65

Now consider,

i.e., P(Y< 50) = P(Y<= 49.5)

By using the continuty correction

P(Y<= 49.5) = P{Z<= (49.5 -64.2)/6.65}

= P( Z <=- 2.21) [since from the normal distribution table]

P(Y<= 49.5) = 0.9864