If and are normally distributed random variables with a mean of 95 and a standard deviation of 20, and that and are independent of each other, then P( and

Question

If  and  are normally distributed random variables with a mean of 95 and a standard deviation of 20, and that  and  are independent of each other, then P(  and  < 60) is:
A) .0401 
B) .0802 
C) .2115 
D) .0016 
E) .1255

Quizplus · Accepted Answer

We know that 11eb6b9e_0665_ed02_8d9d_e590b9312afd_TB8220_11 and 11eb6b9e_0665_ed03_8d9d_291168c1fb6f_TB8220_11 are independent, so we can treat them as two separate normally distributed random variables.

First, we need to find the z-score for a value of 60, using the formula:

z = (x - μ) / σ

z = (60 - 95) / 20

z = -1.75

Next, we use a standard normal distribution table to find the probability of a z-score less than -1.75.

This probability is 0.0401, which is equivalent to P( 11eb6b9e_0665_ed02_8d9d_e590b9312afd_TB8220_11 < 60).

Since 11eb6b9e_0665_ed03_8d9d_291168c1fb6f_TB8220_11 is also normally distributed with a mean of 95 and a standard deviation of 20, we can use the same process to find the probability of 11eb6b9e_0665_ed03_8d9d_291168c1fb6f_TB8220_11 < 60. This probability is also 0.0401.

Because the two events are independent, we can multiply their probabilities to find the probability that they both occur:

P(11eb6b9e_0665_ed02_8d9d_e590b9312afd_TB8220_11 and 11eb6b9e_0665_ed03_8d9d_291168c1fb6f_TB8220_11 < 60) = P( 11eb6b9e_0665_ed02_8d9d_e590b9312afd_TB8220_11 < 60) * P(11eb6b9e_0665_ed03_8d9d_291168c1fb6f_TB8220_11 < 60)

P( 11eb6b9e_0665_ed02_8d9d_e590b9312afd_TB8220_11 and 11eb6b9e_0665_ed03_8d9d_291168c1fb6f_TB8220_11 < 60) = 0.0401 * 0.0401

P( 11eb6b9e_0665_ed02_8d9d_e590b9312afd_TB8220_11 and 11eb6b9e_0665_ed03_8d9d_291168c1fb6f_TB8220_11 < 60) = 0.0016

So the answer is D.

If and Are Normally Distributed Random Variables with a Mean