A random sample of size n 79 is taken from a finite populat

A random sample of size n = 79 is taken from a finite population of size N = 674 with mean = 253 and variance ² = 434. Use Table 1.

Is it necessary to apply the finite population correction factor?

Calculate the expected value and the standard error of the sample mean. (Round “expected value” to a whole number and \"standard error\" to 4 decimal places.)

What is the probability that the sample mean is less than 241? (Leave no cells blank - be certain to enter \"0\" wherever required. Round your answer to the nearest whole number.)

What is the probability that the sample mean lies between 247 and 262? (Use rounded standard deviation. Round \"z\" value to 2 decimal places and final answer to 4 decimal places.)

Solution

A random sample of size n = 79 is taken from a finite population of size N = 674 with mean = 253 and variance ² = 434. Use Table 1.

a-1.

Is it necessary to apply the finite population correction factor?

79/674 =0.1172 which is > 0.05

Answer : yes

Yes

No

a-2.

Calculate the expected value and the standard error of the sample mean. (Round “expected value” to a whole number and \"standard error\" to 4 decimal places.)

fpc = sqrt [ (N - n) / (N - 1) ] =sqrt( 674-79)/674-1)) =0.9403

SE = [ (standard deviation)/sqrt(n) ] * fpc = sqrt(434/79)*0.9403 =2.2039

  Expected value   253

  Standard error    2.2039

b.

What is the probability that the sample mean is less than 241? (Leave no cells blank - be certain to enter \"0\" wherever required. Round your answer to the nearest whole number.)

  Probability = 0

Z value for 241, z=( 241-253)/2.2039 = -5.44

P( x < 241)= P( z < -5.44) = 0.00

c.

What is the probability that the sample mean lies between 247 and 262? (Use rounded standard deviation. Round \"z\" value to 2 decimal places and final answer to 4 decimal places.)

Z value for 247, z=( 247-253)/2.2039 = -2.72

Z value for 262, z=( 262-253)/2.2039 = 4.08

P( 247<x<262) = P( -2.72 <z<4.08)

= p( z < 4.08) – P( z <-2.72) = 1.0000 – 0.0033 = 0.9967

  Probability=0.9967