Question

(a) give the mean and the standard deviation of the sampling distribution of the sample mean x. mean 105 correct: your answer is correct. standard deviation 2.4 correct: your answer is correct. (b) find the probability that x exceeds 112. (round your answer to four decimal places.)

Answer 1

(a) is already answered.

(b) The random variable X is normally distributed with mean μ=105 and standard deviation σ =2.4. Use the transformation $\frac{X-\mu}{\sigma}$ for calculating the probability. This value is called Z-score.

The Z score is $Z=\frac{112-105}{2.4} =2.9167$.

Refer to the standard normal distribution table.

The probability

$P(X>112)=P(Z>2.9167)=1-P(Z<2.9167)=0.0018

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