Question

The average weight of a professional football player in 2009 was 254.9 pounds. Assume the population standard deviation is 40 pounds. A random sample of 32 professional football players was selected. Complete parts a through e.

a. Calculate the standard error of the mean.

​(Round to two decimal places as​ needed.)

b. What is the probability that the sample mean will be less than 238 ​pounds?

​(Round to four decimal places as​ needed.)

c. What is the probability that the sample mean will be more than 235 ​pounds?

​(Round to four decimal places as​ needed.)

d. What is the probability that the sample mean will be between 263 pounds and 268 ​pounds?

​(Round to four decimal places as​ needed.)

e. Identify the symmetrical interval that includes​ 95% of the sample means if the true population mean is 254.9 pounds.

​(Round to the nearest pound as​ needed.)

Answer 1

Solution :

a.

= / n = 40 / 32 = 7.0711

b.

P( < 238) = P(( - ) / < (238 - 254.9) / 7.0711)

= P(z < -2.39)

= 0.0084

Probability = 0.0084

c.

P( > 235) = 1 - P( < 235)

= 1 - P[( - ) / < (235 - 254.9) / 7.0711]

= 1 - P(z < -2.81)

= 1 - 0.0025

= 0.9975

Probability = 0.9975

d.

P(263 < x < 268) = P[(263 - 254.9)/ 7.0711) < (x - ) /  < (268 - 254.9) / 7.0711) ]

= P(1.15 < z < 1.85)

= P(z < 1.85) - P(z < 1.15)

= 0.9678 - 0.8749

= 0.0929

Probability = 0.0929

e.

Middle 95% as the to z values are -1.96 and 1.96

Using z-score formula,  

= z * +

= -1.96 * 7.0711 + 254.9 = 241

and

= 1.96 * 7.0711 + 254.9 = 269

(241 pound and 269 pound)