Question

The united states department of agriculture (usda) found that the proportion of young adults ages 20–39 who regularly skip eating breakfast is 0.238. suppose that lance, a nutritionist, surveys the dietary habits of a random sample of size n=500 of young adults ages 20–39 in the united states.

Answer 1

Answer:

probability  = 0.3557

Explanation:

given data

young adults ages =  20 to 39

skip eating breakfast p = 0.238

random sample of size n = 500

to find out

we find the probability that the number of individuals in Lance's sample who regularly skip breakfast is greater than 122

solution

we use here Normal Approximation to Binomial Distribution

so first consider random variable = x

so

x~ Bin (n,p)   .............1

and here Normal Approximation will be

x~ Normal Approx (np, npq)    .................2

so it will be

x~ (500, 0.238)  

as here we know q will be

q = 1 - p

q = 1 - 0.238

q = 0.762    .............3

so

here x~ Normal Approx (119, 90.678)

and now we get P(X > 122)

so

We will convert it to Z by as that

z = $\frac{x-\mu}{\sigma}$     ................4

and here

mean  $\mu$ = np

and standard deviation $\sigma = \sqrt{npq}$

so here for P(X > 122)

$P(\frac{X-\mu}{\sigma}>\frac{122-119}{\sqrt{90.678}})$     ............5

and it is  P(Z>0.37)

so

probability  = 1 - P(Z<0.37)

now we use here z table for value

probability  = 1-0.6443

probability  = 0.3557