Question

1. The number of pizzas consumed per month by university students is Normally distributed with a mean of 12 and a standard deviation of 5 A. What proportion of students consume more than 14 pizzas per month? B. What is the probability that, in a random sample of 8 students, a sample average of more than 10 pizzas are consumed?

Answer 1

1.

X: Number of pizzas consumed

X follows normal distribution with mean 12 and standard deviation 5

A.

Proportion of students consume more than 14 pizzas per month = P(X>14)

P(X>14) = 1 - P(X $\small \leq$ 14)

Z-score for 14 = (14-12)/5 = 2/5 = 0.4

from standard normal tables, P(Z$\small \leq$0.4) = 0.6554

P(X $\small \leq$ 14)=P(Z$\small \leq$0.4) = 0.6554

P(X>14) = 1 - P(X $\small \leq$ 14) = 1 - 0.6554 = 0.3446

Proportion of students consume more than 14 pizzas per month = 0.3446

B.

If X follows a normal distribution with mean $\mu$ and standard deviation $\sigma$ , then by central limit theorem, sampling distribution of sample mean(sample size:n) follows normal distribution with mean $\mu$ and standard deviation: $\sigma /\sqrt{n}$

Therefore,

$\overline{x}$ :sample Average number of pizzas consumed by 8 students : follow normal distribution with mean 12 and standard deviation = $5/\sqrt{8}$ = 1.7678

Probability that , in a random sample of 8 students, a sample average of more than 10 pizzas are consumed = P($\overline{x}$ > 10)

P($\overline{x}$ > 10) = 1-P()

I < 10

Z-score of 10 =(10-12)/1.7678 = -1.13

from standard normal tables, P(Z$\leq$-1.13) = 0.1292

P(
I < 10
) = P(Z$\leq$-1.13) = 0.1292

P($\overline{x}$ > 10) = 1-P() = 1-0.1292=0.8708

I < 10

Probability that , in a random sample of 8 students, a sample average of more than 10 pizzas are consumed = 0.8708