Answer: 0.7471
Step-by-step explanation:
Given : The Intelligence Quotient (IQ) test scores for adults are normally distributed with a population mean of [tex]\mu=100[/tex] and a population standard deviation of [tex]\sigma=15[/tex].
Sample size = 50
Using formula [tex]z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex], the z-value corresponding to x=98 will be:-
[tex]z=\dfrac{98-100}{\dfrac{15}{\sqrt{50}}}\approx-0.94[/tex]
Z-value corresponding to x=103 will be:-
[tex]z=\dfrac{103-100}{\dfrac{15}{\sqrt{50}}}\approx1.41[/tex]
Using the standard normal distribution table for z-scores, we have
P-value = [tex]P(98<x<103)=P(-0.94<z<1.41)\\\\=P(z<1.41)-P(z<-0.94)\\\\=0.9207301-(1-P(z<0.94))\\\\=0.9207301-(1-0.8263912)=0.9207301-0.1736088=0.7471213\approx0.7471[/tex]
Hence, the probability we could select a sample of 50 adults and find that the mean of this sample is between 98 and 103 = 0.7471
