Question

The average weight of a professional football player in 2009 was pounds. Assume the population standard deviation is pounds. A random sample of professional football players was selected.

Required:
a. Calculate the standard error of the mean.
b. What is the probability that the sample mean will be less than 230 pounds?
c. What is the probability that the sample mean will be more than 231 pounds?
d. What is the probability that the sample mean will be between 248 pounds and 255 pounds?

Answer 1

Answer:

6.286;

0.0165

0.976

0.1995

Step-by-step explanation:

Given that :

Mean, μ = 243. 4

Standard deviation, σ = 35

Sample size, n = 31

1.)

Standard Error

S. E = σ / √n = 35/√31 = 6.286

2.)

P(x < 230) ;

Z = (x - μ) / S.E

P(Z < (230 - 243.4) / 6.286))

P(Z < - 2.132) = 0.0165

3.)

P(x > 231)

P(Z > (231 - 243.4) / 6.286))

P(Z > - 1.973) = 0.976 (area to the right)

4)

P(x < 248)

P(Z < (248 - 243.4) / 6.286))

P(Z < 0.732) = 0.7679

P(x < 255)

P(Z < (255 - 243.4) / 6.286))

P(Z < 1.845) = 0.9674

0.9674 - 0.7679 = 0.1995