# IQ Scores are normally distributed with a mean of 110 and a standard deviation of 15. Find the probabilities for the given values. Round to four decimal places. Probability that if 9 people...

IQ Scores are normally distributed with a mean of 110 and a standard deviation of 15.

Find the probabilities for the given values. Round to four decimal places. Probability that if 9 people are randomly selected, the score is above 127. Find the critical z (alpha divided by two) values if the confidence level is 99%

*print*Print*list*Cite

Given that the IQ scores are normally distributed with mean score 110 and standard deviation. ( people are randomly selected and it is required to find the probabilty that their is above 127.

We know that the sample mean of random sample of size n from a parent normal population with mean M and standard deviation s is also normal with M and standard s/ sqrtn.

So for a random sample of 9 people , the sample mean has the distribution N(M, s/sqrtn) = N(110 scoere , 15/ sqrt9) = N(110 , 15/9) = N(110 , 5).

Thus the sample of 9 people has the IQ score with normal distribution with mean 110 and a standard deviation of 5.

Therefore the mean IQ z = (x -110)/5 is a standard normal variate. So the probabibility that the IQ z > = (127-110)/5 = P{z > = (127-110)/5} = P{z>= 3.4) = 1- P(Z< = 3.4) = 1- 0.9988 = 0.0012. from the tables at http://www.enotes.com/topic/Standard_normal_table.

To find the critical value x for z , such that P( z > x) = 0.005. x = 2.575 from tables.

The 99 % confidence interval for x is given by P {(x - 110|/5) < < 2.317} = 0.99. So x < 110+2.32*5 = 121.59 approximately.