Question

You have information to suggest that a certain continuous variable of a population has a mean of μ=12.44μ=12.44 and a standard deviation of σ=6.78σ=6.78. You are to randomly pick n=68n=68 individuals from this population and observe the value of the population variable on each . This value is to measured as XiXi.

After the random sample of n=68n=68 has been taken, you are asked to consider the behavior of the statistic X¯¯¯¯X¯.

(a) Complete the statement below. Use at least two decimals in each numeric answer.

The distribution of X¯¯¯¯X¯  ? is approximately Normal is exactly Normal is skewed to the right has an unknown shape  with a mean μX¯¯¯¯¯=μX¯=

and a standard deviation σX¯¯¯¯¯=σX¯=

(b) Find the probability that mean of the sample of n=68n=68 is between 11.94 and 12.84. Use at least three decimals in your zz-values, and enter your answer to four decimal

Answer 1

(a) Since we do not know if the population from which the sample is selected, but also that n > 30, therefore

The distribution of X is approximately normal with a mean = 12.44 and a standard deviation = 6.78 / sqrt(68) = 0.82

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(b) To find P (11.94 < X < 12.84) = P(X < 12.84) – P(X < 11.94)

For P( X < 12.84)

Z = (12.84 – 12.44) / 0.82 = 0.488

The probability for P(X < a) from the normal distribution tables is = 0.6872

For P( X < 11.94)

Z = (11.94 – 12.44) / 0.82 = -0.610

The probability for P(X < 11.94) from the normal distribution tables is = 0.2709

Therefore the required probability is 0.6872 – 0.2709 = 0.4163

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