Question

1.)

A distribution of values is normal with a mean of 200 and a standard deviation of 27. From this distribution, you are drawing samples of size 30.

Find the interval containing the middle-most 54% of sample means:

Enter your answer using interval notation. In this context, either inclusive [] or exclusive () intervals are acceptable. Your numbers should be accurate to 1 decimal places.

2.)

A fitness company is building a 20-story high-rise. Architects building the high-rise know that women working for the company have weights that are normally distributed with a mean of 143 lb and a standard deviation of 29 lb, and men working for the company have weights that are normally distributed with a mean of 165 lb and a standard deviation or 33 lb. You need to design an elevator that will safely carry 16 people. Assuming a worst case scenario of 16 male passengers, find the maximum total allowable weight if we want a 0.975 probability that this maximum will not be exceeded when 16 males are randomly selected.

maximum weight = -lb (Round to the nearest pound.)

3.)

A population of values has a normal distribution with μ=188.6and σ=18.4 You intend to draw a random sample of size n=193

Find the probability that a single randomly selected value is less than 188.1.
P(X < 188.1) =

Find the probability that a sample of size n=193 is randomly selected with a mean less than 188.1.
P(x¯ < 188.1) =

Enter your answers as numbers accurate to 4 decimal places.

4.)

Scores for a common standardized college aptitude test are normally distributed with a mean of 501 and a standard deviation of 98. Randomly selected men are given a Test Prepartion Course before taking this test. Assume, for sake of argument, that the test has no effect.

If 1 of the men is randomly selected, find the probability that his score is at least 546.2.
P(X > 546.2) =  Round to 4 decimal places.

If 17 of the men are randomly selected, find the probability that their mean score is at least 546.2.
P(X¯ > 546.2) =  Round to 4 decimal places.

If the random sample of 17 men does result in a mean score of 546.2, is there strong evidence to support the claim that the course is actually effective?

• Yes. The probability indicates that is is (highly ?) unlikely that by chance, a randomly selected group of students would get a mean as high as 546.2.
• No. The probability indicates that is is possible by chance alone to randomly select a group of students with a mean as high as 546.2.

Answer 1

1) Here we have n = 30, = 200 , = 27

X ~ N ( , ² )

Here total area under the normal curve is 1. We have middle area 0.54 ( area between a and b ). So remaining area is 1- 0.54 = 0.46.

Due to symmetry of normal curve area below a is 0.23 and above b is 0.23.

Here we need to find values of a and b.

p( Z < z ) = 0.23

( We use excel formula "=norm.s.inv(0.23) " )

So we get,

z = -0.74

a = 196.35

Now p ( Z > b) = 0.23

1 - p( Z b) = 0.23

p( Z b) = 0.77

( We use excel formula "=norm.s.inv(0.77) " )

So we get,

z = 0.74

b = 203.65

So the interval is ( 196.35, 203.65 ) .