Confusing Sampling Distribution Problem

Ok, I was understanding how to work sampling distribution problems until I came to this one, and I cannot figure out how to do it for the life of me:

Let X be the lifetime of an electronic device. It is known that the average lifetime of the device is 717 days and the standard deviation is 90 days. Let xbar be the sample mean of the lifetimes of 169 devices. The distribution of X is unknown, however, the distribution of xbar should be approximately normal according to the Central Limit Theorem. Calculate the following probabilities using the normal approximation.

(a) P(xbar≤707)=

Mainly, I've been playing around with pnorm with no success. Any suggestions are deeply appreciated.


New Member
You first have to find out where a mean lifetime of 707 falls on the sampling distribution of the sample mean (with n = 169) in terms of standard error units. You have all the necessary information to do so (i.e. the population mean, population standard deviation, sample size.) Does this help get you on the right track or is something else tripping you up?


New Member
You said you've worked some sampling distribution problems, so I'll assume you've seen the formula for the standard error (i.e. sigma/sqrt(n)). In this case your sigma is 90, and n is the amount of scores that make up a sample mean, which is 169. From this information you can calculate the standard error, and then you have to find out where 707 lies in relation to the mean of 717 on the sampling distribution using the standard error. After this it's a matter of reading a z-table to find the probability that a sample falls below or at this value.

If it's still unclear, can you explain exactly what part is giving you trouble?


Less is more. Stay pure. Stay poor.
Not always on the right page, but why would you use the standard error instead of the standard deviation?


New Member
The question is asking for the probability of obtaining a certain sample mean given a sample of 169 scores and so we have to look to the sampling distribution of the mean to find this probability; the spread of means is given by the standard error, not the standard deviation alone.
Thank you for the help so far! I guess the reason I'm still not quite understanding is because we are supposed to use the program R to find answers (I really meant to put that in my original post), and so we don't use Z tables.
Ah! Thank you so much! We had only used pbinom with size and probability, not mean and standard deviation. This clears up so much! Thank you for all of your help!