The distribution of body weights for Caucasian men in a part

The distribution of body weights for Caucasian men in a particular age group has a normal distribution with µ = 190 pounds and = 24 pounds.

1. For randomly selected samples of n = 16 men, what is the expected value of the sampling distribution of the sample mean?

2. For randomly selected samples of n = 16 men, what is the standard error of the sampling distribution of the sample mean?

3. What would be the minimum sample size required to make the standard error of the sampling distribution of the 4. sample mean less than 2.4? (Round to a whole number.)

4. If I randomly select one man, what is the probability that his weight is 195 pounds or more?

5. If I randomly select 100 men, what is the probability that their average weight is 195 pounds or more?

Solution

µ = 190 pounds and = 24 pounds

When n =16 X is N(190,24/rt16) or N (190, 6)

1) Expected value =190

2) Std error = 6

3) Std error = 24/rtn <2.4 means

rt n >10 or n should be atleast 100

4) For one man X is N(190,24)

When x = 195, z = 5/24 = 0.2083

P(Z>0.2083) = 0.5-0.0832 = 0.4168

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5) For n =100, X bar is N(190, 2.4)

So z value for 195 is 2.083

P(Z>2.083) = 0.5-0.4808

= 0.1192