SOLUTION: Please help... 1. A population has a mean of 72 and a standard deviation of 4. In this population, what is the X value corresponding to z = -2.00? I tried -2.00=x-72 x-288

Algebra ->  Probability-and-statistics -> SOLUTION: Please help... 1. A population has a mean of 72 and a standard deviation of 4. In this population, what is the X value corresponding to z = -2.00? I tried -2.00=x-72 x-288       Log On


   

Question 860566: Please help...
1. A population has a mean of 72 and a standard deviation of 4. In this population, what is the X value corresponding to z = -2.00?
I tried -2.00=x-72 x-288
---- = ----- = -36 Is this correct?
4 -8
2. What is the probability of randomly selecting a z-score less than z= -.40 from a normal distribution?
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
 

Hi,

Important to Understand z -values as they relate to the Standard Normal curve:

Below:  z = 0, z = ± 1, z= ±2 , z= ±3 are plotted.  

Note: z = 0 (x value the mean) 50% of the area under the curve is to the left and %50 to the right



1) mean of 72 and a standard deviation of 4.

*Need to KNOw: z+=blue+%28x+-+mu%29%2Fblue%28sigma%29%29%29

    z = (X-72)/4 = -2.00 

       X = -8 + 72 = 64   (z-score tells us X is less than the average of 72

2)P(z = -.40) = 34.46%  = invNorm(-.40)

TI syntax is invNorm(area to left of desired z).