Question

Given the normally distributed random variable X with mean 18 and variance 6.25,
find
(a) P(16 < X < 22).
(b) the value of k1 such that P(X < k1) = 0.2946
(c) the value of k2 such that P(X > k2) = 0.8186
(d) the two cut-off values that contain the middle 90% of the normal curve area

Answer 1

This is a normal distribution question with

a) P(16.0 < x < 22.0)=?

This implies that
P(16.0 < x < 22.0) = P(-0.8 < z < 1.6) = P(Z < 1.6) - P(Z < -0.8)
P(16.0 < x < 22.0) = 0.945200708300442 - 0.2118553985833967

b) Given in the question
P(X < k1) = 0.2946
This implies that
P(Z < -0.54) = 0.2946
With the help of formula for z, we can say that

k1 = 16.65
c) P(X > k2) = 0.8186
P(X < k2) = 1 - 0.8186 = 0.1814
Given in the question
P(X < k2) = 0.1814
This implies that
P(Z < -0.91) = 0.1814
With the help of formula for z, we can say that

k2 = 15.7249
d) Middle 90% in a normal curle lie between 5% and 95%
Given in the question
P(X < x) = 0.05
This implies that
P(Z < -1.645) = 0.05
With the help of formula for z, we can say that

x = 13.8879
Given in the question
P(X < x) = 0.95
This implies that
P(Z < 1.645) = 0.95
With the help of formula for z, we can say that

x = 22.1121
PS: you have to refer z score table to find the final probabilities.
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