Question

The capacity of a lift is 10 people or 1680 lbs. The capacity will be exceeded if 10 people have weights with a mean greater than 1680/10=168 lbs. Suppose the people have weights that are normally distributed with a mean of 171 lbs and a standard deviation of 33 lbs.
a. find probability that if a person is randomly selected, his weight will be greater than 168 lbs.
b. find probability tat 10 randomly selected people will have a mean that is greater than 168 lbs.

Answer 1

Do you know how to find the z-score to solve a ?

Answer 2

to find the z-score, i just look on the z score table right?

Answer 3

it's suppose to be (168-171)/33 and that equals -.0909090909

Answer 4

\[z=\frac{X-\mu}{\sigma}\]

Answer 5

I got that part, but now, am i supposed to look for -.090909 in the table, or on the outside of the table or am i supposed to subtract that from 1. thats where i dont get it.

Answer 6

I got the answer wrong and it gives me .5362 as the solution and I have no clue how they got that. I looked on the score table to see what that fell under but I dont understand how they got this.

Answer 7

In this case drawing a diagram (as follows) should clarify how to use the z-score. The shaded area in my drawing gives the required probability. Standardised normal distribution tables vary unfortunately. I used a table giving only positive values of z from 0 to 3.49. Looking up the probability for z = 0.091, I get p = 0.5363. Effectively the shaded area is the mirror image of that in my drawing, however the value for p is valid.