Question #189334

Let ܺ denote the number of scores on a test. If ܺ is normally

distributed with mean 100 and variance 225, find the probability

that ܺ does not exceed 130.


1
Expert's answer
2021-05-07T11:58:40-0400

σ\sigma = σ2\sqrt{\sigma^2} = 225\sqrt{225} =15

The probability that X < 30 is equal to the blue area under the curve.



Since μ=100\mu=100 and σ=15\sigma=15 we have:

P(X<130) = P(X-μ\mu < 130-100) = P(Xμσ<13010015)(\frac{X-\mu}{\sigma}<\frac{130-100}{15})

Since Xμσ=Z\frac{X-\mu}{\sigma} = Z and 13010015=2\frac{130-100}{15} = 2  we have:

P(X<130) = P(Z<2)

Use the standard normal table



 to conclude that:

P(Z<2) = 0.9772

Answer: P = 0.9772.

Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS