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%

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neela | High School Teacher | (Level 3) Valedictorian

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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.

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