statistics question dealing with probability...?

Question

Choose a person at random and give him an IQ test. The result is a random variable Y. The probability distribution of Y is the Normal distribution with mean μ= 100 and standard deviation σ = 15. The probability P (Y > 120) that the person chosen has IQ score higher than 120 is about?

I tried this a few times, but got different answers, mostly 0.092 and 0.908

Could someone explain the right way to answer this question, and what answer you got, please?

Answer 1

Y ~ N(100, 15^2)

First you have to standardize Y
Z = (Y - 100)/15 ~ N(0,1)
So P(Y>120) = P(Z > [120-100]/15)
P(Z > 1.33)
= 1 - 0.9082 = 0.0918

It helps to draw a sketch showing the area under the normal curve that you're interested in.

Answer 2

Dr. D is correct.

I thought I'd add a comment because the other answer that suggest subtracting 1 to get the degrees of freedom is incorrect. That minus one correction, to get an unbiased estimate, is used when sampling a population.

In your case, you are not sampling a population (i.e., you are not taking some subset n < N of the population and calculating a sample standard deviation and mean based on the n data). You are using the population's mean (100) and standard deviation(15). So no correction of minus one is needed.

Also here's a little trick for a sanity check on your answers.

P(-1sd < x < 1sd) ~ .68 so that P(x > 1sd) ~ .16
P(-2sd < x < 2sd) ~ .95 so that P(x > 2sd) ~ .025
P(-3sd < x < 3sd) ~ .999 so that P(x > 3sd) ~ .001 (or 0)

In your case x > 120-100/15 = 1.33sd; so your answer has to be between .16 and .025 from the above. Clearly .908 is incorrect. If you memorize the three standard deviation (sd) probabilities above, you'll find them handy for checking your answers as you go.

Answer 3

You need to know the population, normally notated with n=, use that when plotting your "z-score" on a Z table.

use the equation Z=(Xbar - mu)/std dev

120-100/15 = 1.33

then look on a z-table and take your sample population number (n) subtract 1. That will give you a degrees of freedom (df). Plot that against your z-score of 1.33 and you get a probability around .0912

Answer 4

Ok, so what you have to do is find the probability that zero children will have this disease first. The probability for each child is an independent, fixed chance of 90%. So take .90 to the fourth power (.90)^4 = .6561 So if that's the chance of having zero kids with a disease, then the chance of having one or more with the disease is 100% - 65.61% or 1-.6561, which gives your answer.

Answer 5

First you have to standardize Y
Z = (Y - 100)/15 ~ N(0,1)
So P(Y>120) = P(Z > [120-100]/15)
P(Z > 1.33)
= 1 - 0.9082 = 0.0918


ARE YOU REALLLY PHYSIC?? (will i ever meet jason dolley?)