I need help with probability statistics!!

Question

The lengths of births in a small rural village are normally distributed with a mean of 269 days and a standard deviation of 15 days.

In what range would you expect to find the middle 98% of most births? 

Between  and .

If you were to draw samples of size 54 from this population, in what range would you expect to find the middle 98% of most averages for the lengths of births in the sample? 

Between  and .

Answer 1

For the normal distribution, 0.02 = 2% of the values are expected to be beyond plus or minus (+-) 2.32 standard deviations above and below the mean.

z = |value - mean| / SD < 2.32 has 100%-2% = 98% of the values.

(+-) (2.32) (SD) = (+-) (2.32) (15) = (+-) 35

So 98% should be between 269-35 = 234 and 269+35 = 304

For means made up of n samples, the distribution is often approximately normal,

with a standard error of the mean, SE, equal to the standard deviation divided by the square root of n.

Here, SD = 15,

SE = 15 / sqrt(54) = 2.0

We would expect the true mean to be within +- 2.34 SE = +- 4.7 in 98% of such samples of 54.

In other words, between about 269-5 = 264 and 269+5 = 274. Means are more tightly distributed than the values from which they are calculated.