A manufacturer produces 25-pound lifting weights. The lowest actual weight is 24 pounds, and the highest is 26 pounds. Each weight is equally likely so the distribution of weights is uniform. A sample of 100 weights is taken.

a. What is the distribution for the weights of one 25-pound lifting weight? What is the mean and standard deviation?

b. What is the distribution for the mean weight of 100 25-pound lifting weights?

c. Find the probability that the mean actual weight for the 100 weights is less than 24.9.

a) Mean = 25, S. d = 0.5774

b) X~N (25, 0.05774)

c) P (X < 24.9) = 0.0416

Step-by-step explanation:

Given:

- Limits of a uniform distribution are:

U (24, 26)

- Sample size n = 100

Find:

a. What is the distribution for the weights of one 25-pound lifting weight? What is the mean and standard deviation?

Solution

The mean and standard deviation value is:

E (X) = (24 + 26) / 2 = 25

Var (X) = (25 - 24) / 3 = 1/3

s. d (X) = sqrt (1/3) = 0.5774

Find:

b. What is the distribution for the mean weight of 100 25-pound lifting weights?

Solution

The random variable X follows a normal distribution:

X~N (25, s. d/100)

X~N (25, 0.05774)

Find:

c. Find the probability that the mean actual weight for the 100 weights is less than 24.9.

Solution

The probability of P (X < 24.9) is:

P (X < 24.9) = P (Z < (24.9-25) / 0.05774)

= P (Z < - 1.7319)

= 0.0416