# How do I solve this probability problem?

Here is the full question:

The lengths of all machine parts made by a company have a distribution that is skewed to the right with a mean of 68 mm and a standard deviation of 4 mm. Find the probability that the mean length of a random sample of 100 parts produced by this company would be: (a) Less than 66.8 mm. (b) Between 66.5mm and 67.7mm. (c) Within 0.6mm of the populations mean. (d) Lower than the population mean by 0.5mm or more.

I answered parts (a) and (b), and got them to be 0.13% and 22.65% respectively. I just don't get what parts (c) and (d) mean. In part (c), I don't get what 'within' implies, and in part (d) I am not sure if I am supposed to decrease by 0.5mm then solve it as in part (a) or not.

• which distrivution did you choose to use? i know the question mentions a skew specifically, but the data appears to be continuous, so did you use the normal distribution? Oct 9, 2017 at 17:56
• yes, I choose the normal distribution. Oct 9, 2017 at 18:01

In spite of the skewness of the population, we hope that the Central Limit Theorem guarantees that sample mean of $n = 100$ observations is approximately distributed as $\mathsf{Norm}(\mu = 68, \sigma = 4/\sqrt{100} = 0.4).$

If you have access to statistical software or a statistical calculator you can find the desired probabilities without "standardizing," which is necessary if you use printed normal tables. For example, in R statistical software the normal CDF is pnorm with appropriate parameters to show $\mu$ and $\sigma.$

For example, to check your answer to part (a): $P(\bar X < 66.8) = .00135.$

 pnorm(66.8, 68, .4)
## 0.001349898


For part (c), you want $P(68-.6 < \bar X < 68 + .6) = P(67.4 < \bar X < 68.6) = 0.8664.$

 pnorm(68.6, 68, .4) - pnorm(67.4, 68, .4)
## 0.8663856


In the sketch below, you want the area under the normal curve between the vertical broken red lines. By hand, you can't make such an accurate sketch, but you should always try to make a rough sketch to keep you from making needless errors. This is, indeed, the same as the probability that a standard normal random variable is within 1.5 standard deviations of its mean 0. Using printed tables, you should be able to get very close to this answer. In R, it looks like this (no extra parameters are needed for the standard normal distribution):

pnorm(1.5)-pnorm(-1.5)
## 0.8663856


For part (d), you want $P(\bar X < 68-.05 = 67.5):$

pnorm(67.5, 68, .4)
## 0.1056498


Of course, you should use whatever method your course requires, but I hope you will come close to the numbers I have given. Also, you should understand that the mean of a sample from a skewed population distribution is only approximately normal. Unless the population is extremely skewed, I would trust the first two decimal places to be correct.

• I understood part d, but, in part c, I still don't understand what 'within 0.6mm of the population mean' imply. Is it the same as saying 'Find the probability that the mean length of a random sample of 100 parts produced by this company would be 0.6mm more than the population mean and 0.6mm less than the population mean' ? Oct 10, 2017 at 2:47
• Pop mean is $\mu = 68.$ "Within $.6$ of pop mean" is a standard expression for "In the interval $68 \pm .6.$" or "In the interval $(68-.6, 68+.6).$" I don't see how it would mean anything else. Oct 10, 2017 at 3:38
• You might want to change P(X < 68 - .05 = 67.5) to P(X < 68 - .5 = 67.5). And thank you. Oct 10, 2017 at 11:08

Part C, I assume, means that you take the 0.6mm it meantions and divide that by the s.d. of 4, to get 0.15 standard deviations from the mean, then use phi on 0.15, to get a probability over 50%

BUT, that probability assumes you want everything under 0.15 s.d. above the mean, but you only want the 0.15 s.d. either side of the mean, so you take 1 - 2*phi(0.15), this removes everything bar the 0.15 s.d. in the middle.

Then part D is similar, 0.5/4 to get 1/8 standard deviations, then 1- phi(1/8), as you are looking exclusively for data below the mean, by 1/8 of a standard deviation.

• Maybe you want to reconsider this using the distribution of the mean of a sample of size 100. Also in part (c), it seems it is the central part that matters.. Oct 9, 2017 at 20:48