IQ scores (as measured by the Stanford-Binet intelligence test) in a certain country are normally distributed with a mean of 95 and a standard deviation of 18. Find the approximate number of people in the country (assuming a total population of 323,000,000) with an IQ higher than 126. (Round your answer to the nearest hundred thousand.)

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Tundexi Tundexi · Answer 1 · 2020-11-25T16:36:11+01:00

Answer:

13,797,339 people

Step-by-step explanation:

Given that μ = 95 and σ = 18

The total number of population is N = 323,000,000

Now, we calculate the probability that IQ is higher than 126

P(X>126) = 1 - P[Z ≤ 126 - 95 / 18]

P(X>126) = 1 - P[Z ≤ 1.72]

P(X>126) = 1 - [ NORMSDIST(1.72)] Using MS Excel

P(X>126) = 1 - 0.957283779

P(X>126) = 0.042716221

Hence, the required probability is 0.042716221

Now, we calculate the approximate number of people in the county with an IQ higher than 126

Number of people = P(X>128) * N

Number of people = 0.042716221 * 323,000,000

Number of people = 13797339.383

Number of people = 13,797,339

Therefore, the approximate number of people in the country with an IQ higher than 126 is 13,797,339 in population

raphealnwobi raphealnwobi · Answer 2 · 2021-12-02T12:41:28+01:00

There are 13800000 people in the country with an IQ higher than 126

The z score is used to determine by how many standard deviations the raw score is above or below the mean. It is given by:

[tex]z=\frac{x-\mu}{\sigma} \\\\where\ x=raw\ score,\mu=mean,\sigma=standard\ deviation[/tex]

Given that μ = 95, σ = 18;

For x > 126:

[tex]z=\frac{126-95}{18} =1.72[/tex]

From the normal distribution table, P(z > 1.72) = 1 - P(z < 1.72) = 1 - 0.9573 = 0.0427

For a total population of 323,000,000:

Number of people in the country = 0.0427 * 323,000,000 = 13800000

Hence there are 13800000 people in the country with an IQ higher than 126

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