Question

Assuming all of the distributions are normal, find the x for each of the cases:

a. P(Z < k) = 0.92

b. P(Z > k) = 0.72

c. P(−1 < Z < k) = 0.60

d. (A-Grade) P(k < Z < 1.7) = 0.57

e. (A-Grade) P(Z = k) = 0.00

Answer 1

Solution(a)
Given in the question
P(Z<k) = 0.92
Here we need to find Z-score for which Area is 0.92
This Z-score can be found from Z-table So
k = 1.405
P(Z<1.405) = 0.92
Solution(b)
P(Z>k) = 0.72
here we need to k
P(Z>k) = 1 - P(Z<=k) = 0.72
P(Z<=k) = 1 - 0.72 = 0.28
From Z table we found Z-score i.e.
k = -0.583
So P(Z>-0.583) = 0.72
Solution(c)
P(-1<Z<k) = 0.60
Here we need to k. this can be written as follows:
P(-1<Z<k) = P(Z<k) - P(Z<-1) = 0.60
P(Z<-1) from Z table is 0.1587
P(Z<k) - P(Z<-1) = 0.60
P(Z<k) -0.1587 = 0.60
P(Z<k) = 0.60 + 0.1587 = 0.7587
from Ztable we found k value
k = 0.702
So P(-1<Z<0.702) = 0.60
Solution(d)
P(k<Z<1.7) = 0.57
Here we need to calculate k-value
Above equation can be written as
P(k<Z<1.7) = P(Z<1.7) - P(Z<k) = 0.57
P(Z<1.7) From z table is 0.9554
P(Z<1.7) - P(Z<k) = 0.57
0.9554 - P(Z<k) = 0.57
P(Z<k) = 0.9554 - 0.57 = 0.3854
So k value from Z table is -0.29
P(-0.29<Z<1.7) = 0.57
Solution(e)
P(Z=k) = 0.00
Here we can accept any value of k because in a normal distribution curve we calculate cumulative probability or area. so any k value area will be zero.