IQ scores are normally distributed with a mean of 100 and a standard deviation of 15 . Out of a randomly selected 1350 people from the population, how many of them would have an IQ between 95 and 123 , to the nearest whole number?
Q. IQ scores are normally distributed with a mean of 100 and a standard deviation of 15 . Out of a randomly selected 1350 people from the population, how many of them would have an IQ between 95 and 123 , to the nearest whole number?
Identify Mean and Standard Deviation: Identify the mean and standard deviation of the IQ scores.Mean (μ) = 100Standard deviation (σ) = 15
Calculate Z-Scores: Calculate the z-scores for the IQ scores of 95 and 123. The z-score formula is z=σX−μ, where X is the value from the data set. For IQ = 95: z=1595−100=15−5=−31≈−0.33 For IQ = 123: z=15123−100=1523≈1.53
Find Area Under the Curve: Use the standard normal distribution table or a calculator to find the area under the curve between the z-scores of −0.33 and 1.53. The area under the curve from the mean to z=−0.33 is approximately 0.3707. The area under the curve from the mean to z=1.53 is approximately 0.9370.
Calculate Area Between Z-Scores: Calculate the area between the two z-scores by subtracting the area from the mean to z=−0.33 from the area from the mean to z=1.53.Area between z=−0.33 and z=1.53 = 0.9370−0.3707=0.5663
Find Number of People: Multiply the area by the total number of people to find the number of people with an IQ between 95 and 123. Number of people = Total population × Area between z-scores Number of people =1350×0.5663≈764.805
Round to Nearest Whole Number: Round the result to the nearest whole number, as we cannot have a fraction of a person.Number of people ≈765