The average weight of a National Football League (NFL) player in 2009 is 252.8 p

Question

The average weight of a National Football League (NFL) player in 2009 is 252.8 pounds. Assume the population standard deviation is 25 pounds. A random sample of 38 NFL players was selected.

calculate standard error of mean

what is the probability that the sample mean will be less than 246 pounds

what is the probability that the sample mean will be more than 249 pounds

what is the probability that the sample mean will be between 254 and 258 pounds

identifythe symmetrical interval that includes 95% of the sample means if the true population mean is 252.8 pounds

Explanation / Answer

Normal Distribution
Mean ( u ) =252.8
Standard Deviation ( sd )=25
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
a)
P(X < 246) = (246-252.8)/25
= -6.8/25= -0.272
= P ( Z <-0.272) From Standard Normal Table
= 0.3928                  
b)
P(X > 249) = (249-252.8)/25
= -3.8/25 = -0.152
= P ( Z >-0.152) From Standard Normal Table
= 0.5604                  
c)
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 254) = (254-252.8)/25
= 1.2/25 = 0.048
= P ( Z <0.048) From Standard Normal Table
= 0.51914
P(X < 258) = (258-252.8)/25
= 5.2/25 = 0.208
= P ( Z <0.208) From Standard Normal Table
= 0.58239
P(254 < X < 258) = 0.58239-0.51914 = 0.0632                  
d)
P ( Z < x ) = 0.95
Value of z to the cumulative probability of 0.95 from normal table is 1.645
P( x-u/s.d < x - 252.8/25 ) = 0.95
That is, ( x - 252.8/25 ) = 1.64
--> x = 1.64 * 25 + 252.8 = 293.925                  

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