Question

The average weight of a professional football player in 2009 was 245.3 pounds. Assume the population standard deviation is 40 pounds. A random sample of 35 professional football players was selected. Complete parts a through e a. Calculate the standard error of the mean. = 6.76 (Round to two decimal places as needed.) b. What is the probability that the sample mean will be less than 236 pounds? P( <236) 0845 (Round to four decimal places as needed.) c. What is the probability that the sample mean will be more than 239 pounds? P (x>239) 8238 Round to four decimal places as needed.) d. What is the probability that the sample mean will be between 250 pounds and 257 pounds? P (250 SXS 257)= .2002 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 2453 pounds (Round to the nearest pound as needed.) Enter your answer in the edit fields and then click Check Answer

Answer 1

Solution :

Given that

$\sigma$_{$\bar x$} = $\sigma$ / $\sqrt$ n = 6.76

Using Empirical rule,

P($\mu$_{$\bar x$}  - 2$\sigma$_{$\bar x$} < $\bar x$ < $\mu$ _{$\bar x$}  + 2$\sigma$_{$\bar x$} ) = 95%

P(245.3 - 2 * 6.76 < $\bar x$ < 245.3 + 2 * 6.76) = 95%

P(232 < $\bar x$ < 259) = 95%

232 $\leq$$\bar x$$\leq$ 259