The Intelligence Quotient (IQ) test scores for adults are normally distributed with a mean of 100 and a population standard deviation of 15. What is the probability we could select a sample of 50 adults and find that the mean of this sample is between 98 and 103? Show your solution.

A.) 0.3264

B.) 0.9428

C.) 0.4702

D.) 0.7471

E.) 0.6531

D.) 0.7471

Step-by-step explanation:

Mean=μ=100

Standard deviation=σ=15

We know that IQ score are normally distributed with mean 100 and standard deviation 15.

n=50

According to central limit theorem, if the population is normally distributed with mean μ and standard deviation σ then the distribution of sample taken from this population will be normally distributed with mean μxbar and standard deviation σxbar=σ/√n.

Mean of sampling distribution=μxbar=μ=100.

Standard deviation of sampling distribution=σxbar=σ/√n=15/√50=2.1213.

We are interested in finding the probability of sample mean between 98 and 103.

P(98<xbar<103)=?

Z-score associated with 98

Z-score=(xbar-μxbar)/σxbar

Z-score=(98-100)/2.1213

Z-score=-2/2.1213

Z-score=-0.94

Z-score associated with 103

Z-score=(xbar-μxbar)/σxbar

Z-score=(103-100)/2.1213

Z-score=3/2.1213

Z-score=1.41

P(98<xbar<103)=P(-0.94<Z<1.41)

P(98<xbar<103)=P(-0.94<Z<0)+P(0<Z<1.41)

P(98<xbar<103)=0.3264+0.4207

P(98<xbar<103)=0.7471

Thus, the probability that the sample mean is between 98 and 103 is 0.7471.

addition

step-by-step explanation: