Could someone solve this, please. I understand it is hypothesistesting, but not

Question

Could someone solve this, please. I understand it is hypothesistesting, but not quite sure how to solve it. Thanks.

Rockwell hardness of pins of a certain type is known to have a meanvalue of 50 and a std. deviation of 1.2.
a) If the distribution is normal, what is the probability that thesample mean hardness for a random sample of 9 pins is at least51.

b) What is the approximate probability that the sample meanhardness for a random sample of 40 pins is at least 51?

Explanation / Answer

The population is known to have a mean = 50 and =1.2 (a) We are taking a sample of size n = 9 The parameters of the sampling distribution of the means thexbar= = 50 and xbar= /n = 1.2/9 = 0.4 We want to know the probability of xbar > 51. Since thedata is normally distributed, we can convert this to a z-score anduse the normal distribution to compute the proportion of means witha hardness above 51. Now we must calculate the z-score. z = (xbar - ) / xbar= (51 - 50)/0.4 =2.5 Now we can use a normal distribution table (or use an onlinecalculator, I used thishttp://davidmlane.com/hyperstat/z_table.html) to convert thez-score into a probability: P(z>2.5) = 0.006210 (b) We are taking a sample of size n = 40 The parameters of the sampling distribution of the means thexbar= = 50 and xbar= /n = 1.2/40 = 0.19 We want to know the probability of xbar > 51. Since thedata is normally distributed, we can convert this to a z-score anduse the normal distribution to compute the proportion of means witha hardness above 51. Now we must calculate the z-score. z = (xbar - ) / xbar= (51 - 50)/0.19 =5.26 5.26 standard devations above the mean is *a lot*. Theprobability of drawing a sample so many standard deviations fromthe mean is extremely low, almost 0. Normal distributiontables rarely go this high. Using an online calculator, I used thishttp://davidmlane.com/hyperstat/z_table.html) to convert thez-score into a probability: P(z>5.26) < 0.0000001

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