Please answer the following question.

Question

Suppose x has a distribution with a mean of 60 and a standard deviation of 21. Random samples of size n = 36

 are drawn.

Find the z value corresponding to x = 53.

z=

Find P(x < 53).

 (Round your answer to four decimal places.)

P(x < 53) =

Answer 1

A z-score corresponding to a sample mean of x = 53 can be computed from the population mean of 60, the population standard deviation of 21, and the size of the samples (in this case 36). The first step is to compute the standard error of the mean, which is equal to the population standard deviation (21) divided by the square root of the sample size. The square root of 36 is 6. Dividing 21 (the population standard deviation) by 6, we get 3.5, which is the standard error of the mean. We can now compute the z-score corresponding to the mean of our sample. We do this by subtracting the population mean, 60, from the sample mean, 53, and then dividing by the standard error of the mean, 3.5. This gives (53-60)/3.5, or -2.0000 (to four decimal places, which is more than would be needed for most purposes). This is the z-score for our sample. The final step is to compute p(x<53), the probability of getting a sample size lower than 53. To do this we consult a table of z-scores and look up the probability of obtaining a z-score of less than -2.00. Since such tables only show positive z-scores, we look up a z-score of +2.00 and check the percentage of scores lying above this value (i.e. in the upper tail of the distribution). The value we find there is 2.28, meaning that 2.28% of scores will be more extreme than our obtained z value of 2.00. This means that p(x<53), the probability of obtaining a sample mean of less than 53, is 2.28%