The IQ scores are normally distributed with a mean of 100 and a standard deviati

Question

The IQ scores are normally distributed with a mean of 100 and a standard deviation of 15.

100 + or - sd (15) =                                                 1 sd =?       2 sds = ? 3sds = ?

2.. What is the probability that a randomly selected person has an IQ between 85 and 115?

3. Find the 90th percentile of the IQ distribution. Hint: Using a normal table, Probability = .4500

Z = 1.645 sd

4. If a random sample of 100 people is selected, what is the standard deviation of the sample mean?

standard deviation = sd/n

Explanation / Answer

1) mean +/- 1SD = (100-15 ; 100+15 ) =(85 ; 115)

mean +/- 2SD =70 ; 130

mean +/- 3SD =55 ; 145

2) pobabilty that a randomly selected person has an IQ between 85 and 115 =P(85<X<115)

=P((85-100)/15<Z<(115-100)/15) =P(-1<Z<1) =0.8413- 0.1587 =0.6827

3) for 90th percentile ; zvalue =1.2815

hence corresponding value =mean +z*std deviation =100+1.2815*15 =119.22

4)std deviation od sample mean =std deviaiton/(n)1/2 =15/(100)1/2 =1.5

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