Suppose a random sample of n = 16 observations is selected from a population that is normally distributed with mean equal to 102 and standard deviation equal to 10.
a) Give the mean and the standard deviation of the sampling
distribution of the sample mean x.
mean =
standard deviation =
b) Find the probability that x exceeds 106. (Round your answer to four decimal places.)
c) Find the probability that the sample mean deviates from the population mean μ = 102 by no more than 4. (Round your answer to four decimal places.)
Solution :
Given that ,
a) _{} = 102
_{} = / n = 10 / 16 = 2.5
b) P( > 106) = 1 - P( < 106)
= 1 - P[( - _{} ) / _{} < (106 - 102) / 2.5]
= 1 - P(z < 1.60)
= 1 - 0.9452
= 0.0548
c) P( < 106 ) = P(( - _{} ) / _{} < (106 - 102) / 2.5)
= P(z < 1.60)
Using z table
= 0.9452
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