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.)

Number of People having IQ higher than 135 = 1,298,460 Nos

Step-by-step explanation:

Approach 1 using Normal Distribution Tables:

As we know that for normal distribution z(x) = (x-Mu)/SD - Equation 1

Given Data:

Mean of IQ scores = Mu = 90

Standard Deviation in IQ scores = SD = 17

Let 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 decimals

Using the Normal distribution tables we have:

P(A) = (P<=135) = 0.99598

As we know that:

Probability of having IQ scores more than 135 = 1 - Probability of having IQ scores less than or equal to 135

Probability of having IQ scores more than 135 = 1 - P(A)

Probability of having IQ scores more than 135 = 1 - 0.99598

Probability of having IQ scores more than 135 = 0.00402

As total population of country is equal to 323,000,000 Nos therefore number of people having IQ scores more than 135 = 0.00402 * 323,000,000

Number of people having IQ scores more than 135 = 1,298,460

Approach 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.

I THINK IT IS 5.8609 I AM SO SORRY IF I AM WRONG

Step-by-step explanation: