The Intelligence Quotient (IQ) test scores for adults are normally distributed with a population 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?

Answer:[tex] P( \bar X >104) = P(Z > \frac{104-100}{\frac{15}{\sqrt{50}}}) = P(Z>1.886)[/tex]And we can use the complement rule and the normal standard distribution or excel and we got:[tex] P(z>1.886) = 1-P(Z<1.886) = 1-0.970 = 0.03[/tex]Step-by-step explanation:Previous concepts Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean". The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  Solution to the problem Let X the random variable that represent the IQ of a population, and for this case we know the distribution for X is given by: [tex]X \sim N(100,15)[/tex]  Where [tex]\mu=100[/tex] and [tex]\sigma=15[/tex] We select a sample of n = 50 and we want to find the probability that:[tex]P(\bar X >104)[/tex]Since the distribution for X is normal then we know that the distribution for the sample mean [tex]\bar X[/tex] is given by: [tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex] And we can use the z score formula given by:[tex] z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}} [/tex]And using this formula we got:[tex] P( \bar X >104) = P(Z > \frac{104-100}{\frac{15}{\sqrt{50}}}) = P(Z>1.886)[/tex]And we can use the complement rule and the normal standard distribution or excel and we got:[tex] P(z>1.886) = 1-P(Z<1.886) = 1-0.970 = 0.03[/tex]