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

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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)   (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 = (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).
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