Dr. Smith wants to use only students with "normal" IQ in his experiment. He defines "normal" as anyone who scores in the middle 50% of IQ scores. Using this rule, what will be the lowest IQ score that could be included in the study and what would be the highest IQ score that could be included in the study? You know that the population of IQ scores are normally distributed, have µ = 100, and have σ = 15. (1 point)

Answer:Lowest IQ: 89.875Highest IQ: 110.125Step-by-step explanation:Problems of normally distributed samples can be solved using the z-score formula.In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:[tex]Z = \frac{X - \mu}{\sigma}[/tex]The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.In this problem, we have that:[tex]\mu = 100, \sigma = 15[/tex].He defines "normal" as anyone who scores in the middle 50% of IQ scores. Using this rule, what will be the lowest IQ score that could be included in the study and what would be the highest IQ score that could be included in the study?The middle 50% is the interval from the 25th percentile to the 75th percentile.Lowest IQ:This is the measure in the 25th percentile. That is X when Z has a pvalue of 0.25. So it is [tex]Z = -0.675[/tex][tex]Z = \frac{X - \mu}{\sigma}[/tex][tex]-0.675 = \frac{X - 100}{15}[/tex][tex]X - 100 = 15*(-0.675)[/tex][tex]X = 89.875[/tex]Highest IQ:This is the measure in the 75th percentile. That is X when Z has a pvalue of 0.75. So it is [tex]Z = 0.675[/tex][tex]Z = \frac{X - \mu}{\sigma}[/tex][tex]0.675 = \frac{X - 100}{15}[/tex][tex]X - 100 = 15*(0.675)[/tex][tex]X = 110.125[/tex]