Question

In the library on a university campus, there is a sign in the elevator that indicates a limit of 16 persons. Furthermore, there is a weight limit of 2500 points. Assume the average weight of persons on campus is 150 pounds, the standard deviation is 27 pounds, and that the distribution of weights is approximately normal. If a random sample of 16 people is taken.
A.What is the expected value of the sample mean of their weights?
B.What is the standard deviation of the sampling distribution?
C.What average weight for these 16 people will result in the total weight exceeding the weight limit of 2500 points?
D. What is the chance that a random sample of 16 persons on the elevator will exceed the weight limit?

Answer 1

Answer:

a) 150 pounds

b) 6.75

c) 156.25

d) 0.177

Step-by-step explanation:

We are given the following information in the question:

Mean, μ =  150 pounds

Standard Deviation, σ = 27 pounds

We are given that the distribution of weight of persons on campus is a bell shaped distribution that is a normal distribution.

a) expected value of the sample mean of their weights

$\bar{x} = \mu = 150\text{ pounds}$

b) standard deviation of the sampling distribution

$\dfrac{\sigma}{\sqrt{n}} = \dfrac{27}{\sqrt{16}} = 6.75$

c)  average weight for these 16 people will result in the total weight exceeding the weight limit of 2500 points

$\dfrac{\text{Limit}}{\text{Number of people}} = \dfrac{2500}{16} = 156.25$

d) P(sample of 16 persons on the elevator will exceed the weight limit)

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

P(x > 156.25)

$P( x > 156.25) = P( z > \displaystyle\frac{156.25 - 150}{6.75}) = P(z > 0.9259)$

$= 1 - P(z \leq 0.9259)$

Calculation the value from standard normal z table, we have,  

$P(x > 610) = 1 - 0.823 = 0.177 = 17.7\%$

0.177 is the probability that a random sample of 16 persons on the elevator will exceed the weight limit.