# What value would you expect to find for the mean of all these sample means? I got 5.23 but thats wrong. And Find the area under the standard normal curve between z=-1.75 and z= 1.22. I got 0.9289A...

What value would you expect to find for the mean of all these sample means? I got 5.23 but thats wrong. And Find the area under the standard normal curve between z=-1.75 and z= 1.22. I got 0.9289

A certain population has a mean of 523 and a standard deviation of 28. Many samples of size 36 are randomly selected and the means calculated.

A. What value would you expect to find for the mean of all these sample means?

I got 5.23 but thats wrong.

B. What value would you expect to find for the standard deviation of all these sample means?

I got .87

Another question

Find the area under the standard normal curve between z=-1.75 and z= 1.22

I got 0.9289

B. Assume a normal distribution, find the z-score associated with the 65th perctile?

C. For a normal distribution of x-values, where u= 45 and a= 5.1, find the value of z when x=61?

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(1) Given `mu=523,sigma=28` with samples of size 36:

(a) **The mean of the samples should be close to 523** -- the central limit theorem states that as the number of samples increases, the mean of the samples approaches the mean of the population.

How close? You can compute a range with a given probability.

(b) **The standard deviation of the samples is approximately** `s=sigma/sqrt(n)=28/sqrt(36)=4.67` , again by the central limit theorem.

(2) Find the area between `z=-1.75` and `z=1.22` :

Using the standard normal table the area to the left of `z=-1.75` is .0401 and the area to the left of `z=1.22` is .8888

**The area between the two z values is .8888-.0401=.8487** (My graphing calculator gives me .8487083856)

** Some standard normal tables list the area differently; instead of to the left of a given z value they mark from 0. Make sure you know how to use the table you are looking at**

(3) Find the z score associated with the 65th percentile:

The area under the standard normal curve is 1. If 65% of the scores lie below a given score, then the z-value will have area to its left of .65. **Looking in a standard normal table for .6500 we find the z-score lies between .38 and .39**. Your instructor may have instructions on how to deal with this -- some go to the nearest value, some use interpolation. My calculator gives .3853204726

(4) Find the normalized score for x=61 when `mu=45,sigma=5.1` :

**The normalized score is the number of standard deviations above or below the mean that a given value is at; the formula is `z=(X-mu)/sigma` so `z=(61-45)/5.1=16/5.1=3.14` **