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

set each factor equal to 0

4x-3 = 0

2x-1 = 0

solve for x.

4x-3=0

4x=3

x= 3/4

solve for x. 

2x-1= 0

2x= 1

x= 1/2

eventually after a few more steps, your answer should be 1/2 is less than or equal to x and 3/4 is greater than or equal to x

answer: about 22.67

step-by-step explanation: