IQ scores (as measured by the Stanford-Binet intelligence test) in a certain country are normally distributed with a mean of 90 and a standard deviation of 17. Find the approximate number of people in the country (assuming a total population of 323,000,000) with an IQ higher than 135. (Round your answer to the nearest hundred thousand.)

Answer:Number of People having IQ higher than 135 = 1,298,460 NosStep-by-step explanation:Approach 1 using Normal Distribution Tables:As we know that for normal distribution z(x) = (x-Mu)/SD - Equation 1Given Data:Mean of IQ scores = Mu = 90Standard Deviation in IQ scores = SD = 17Let we assume that probability of having IQ scores less than or equal 135 = P(A)The z for IQ scores of 135 may be calculated using equation 1 as under:z(135) = (135-90)/17 = 2.6470 = 2.65 //After rounding off to two decimalsUsing the Normal distribution tables we have:P(A) = (P<=135) = 0.99598As we know that:Probability of having IQ scores more than 135 = 1 - Probability of having IQ scores less than or equal to 135Probability of having IQ scores more than 135 = 1 - P(A)Probability of having IQ scores more than 135 = 1 - 0.99598Probability of having IQ scores more than 135 = 0.00402As total population of country is equal to 323,000,000 Nos therefore number of people having IQ scores more than 135 = 0.00402 * 323,000,000Number of people having IQ scores more than 135 = 1,298,460Approach 2 using Excel or Google Sheets:Probability of having IQ scores more than 135 = 1 - norm.dist(135,Mu,SD,Commutative)Probability of having IQ scores more than 135 = 1 - norm.dist(135,90,17,1)And then after multiplying with the total population of country we will get the number of people having IQ scores more than 135.PS: The normal distribution tables are being attached for yours easiness.