SOLUTION: Use the probability distribution for the random variable x to answer the question. x 0 1 2 3 4 5 p(x) 0.35 0.2 0.19 0.17 0.06 0.03 Locate the interval 𝜇 ± 2𝜎 on the

Algebra ->  Probability-and-statistics -> SOLUTION: Use the probability distribution for the random variable x to answer the question. x 0 1 2 3 4 5 p(x) 0.35 0.2 0.19 0.17 0.06 0.03 Locate the interval 𝜇 ± 2𝜎 on the       Log On


   

Question 1206571: Use the probability distribution for the random variable x to answer the question.
x 0 1 2 3 4 5
p(x)
0.35 0.2 0.19 0.17 0.06 0.03
Locate the interval
𝜇 ± 2𝜎
on the x-axis of a probability histogram. You are given that the mean is 1.48 and the standard deviation is 1.42. What is the probability that x will fall into this interval?
Answer by math_tutor2020(3817) About Me  (Show Source):
You can put this solution on YOUR website!

𝜇 = greek letter mu = mean = 1.48
𝜎 = greek letter sigma = standard deviation = 1.42

𝜇 ± 2𝜎 breaks down into 𝜇 - 2𝜎 and 𝜇 + 2𝜎

This will mean 𝜇 - 2𝜎 < x < 𝜇 + 2𝜎

𝜇 - 2𝜎 = 1.48 - 2*1.42 = -1.36
𝜇 + 2𝜎 = 1.48 + 2*1.42 = 4.32

We're focused on the interval -1.36 < x < 4.32

If x is an integer then we focus on the sub-interval -1+%3C=+x+%3C=+4
Furthermore, the smallest x can be in this case is x = 0, so we're really only going to focus on 0+%3C=+x+%3C=+4

Add up the probability values associated with x = 0 through x = 4.
P(0)+P(1)+P(2)+P(4) = 0.35+0.2+0.19+0.17+0.06 = 0.97 which is the final answer

Or as an alternate route
P(5) = 0.03
P(value between 0 and 4) = 1 - P(5) = 1 - 0.03 = 0.97

There's a 97% chance of picking an x value between 0 and 4 inclusive of both endpoints.
This is equivalent to saying there's a 97% chance of picking an x value such that -1.36 < x < 4.32; where each endpoint is two standard deviations away from the mean.


Answer: 0.97