Question

According to the UF sexual health student survey, 19% of US students have never had sex of any kind. suppose we randomly sample 300 students. what is the probability that 50 or more of the students in the sample say they have never had sex

Answer #1

Let x = Number of US students who have never had sex of any kind

sample size = n = 300

proportion of US students who have never had sex of any kind = p = 0.19

Here n * p = 300 * 0.19 = 57

n * q = 300 * ( 1 - 0.19) = 243

Both n*p and n*q both are greater than 10

So, **normal approximation to binomial is appropriate to
use**

**By using normal approximation to binomial**

mean =

SD =

we want to find the probability that 50 or more of the students in the sample say they have never had sex

i.e.

By using continuity correction factor we have to calculate

By using Z score we can calculate this probability

For x = 49.5

i.e. we want to calculate

.................( From Z table)

.................( ANSWER)

**Probability that 50 or more of the students in the
sample say they have never had sex is
0.8653 .............( ANSWER)**

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