Question

IQ scores are normally distributed with a mean of 100 and a standard deviation of 15. Out of a randomly selected 1450 people from the population, how many of them would have an IQ between 106 and 125, to the nearest whole number?

Answer 1

To solve this question, we'll need to use the properties of normal distribution. Specifically, we'll use the z-scores, which tell us how many standard deviations away from the mean a given value is. The formula for calculating a z-score is:

Z = (X - μ) / σ

where:

- X is the value we're interested in,

- μ is the mean, and

- σ is the standard deviation.

The z-score corresponding to an IQ of 106 is (106 - 100) / 15 ≈ 0.40, and the z-score corresponding to an IQ of 125 is (125 - 100) / 15 ≈ 1.67.

Next, we need to determine the proportion of individuals in a normal distribution that falls between these z-scores. This is found by looking up these z-scores in a standard normal distribution table, or using a software function that provides the cumulative distribution function of the standard normal distribution.

The approximate values are:

- The cumulative probability associated with a z-score of 0.40 is about 0.6554.

- The cumulative probability associated with a z-score of 1.67 is about 0.9525.

The proportion of individuals with IQs between 106 and 125 is the difference between these probabilities, or about 0.9525 - 0.6554 = 0.2971.

Now, to find the number of people out of 1450 with IQs between 106 and 125, we simply multiply this proportion by 1450 and round to the nearest whole number:

0.2971 * 1450 ≈ 431

So, we expect about 431 people out of the randomly selected 1450 to have an IQ between 106 and 125.

Answer 2

Explanation:

To find out how many people would have an IQ between 106 and 125, we need to calculate the area under the normal distribution curve between these two IQ values. We can do this by calculating the z-scores corresponding to these IQ values and then using a standard normal distribution table or a calculator.

First, let's calculate the z-score for an IQ of 106 using the formula:

z = (x - μ) / σ

where x is the IQ score (106), μ is the mean (100), and σ is the standard deviation (15).

z = (106 - 100) / 15 = 0.4

Next, let's calculate the z-score for an IQ of 125:

z = (x - μ) / σz = (125 - 100) / 15 = 1.67

Using a standard normal distribution table or a calculator, we can find the corresponding cumulative probabilities for these z-scores.

The cumulative probability for a z-score of 0.4 is approximately 0.6554.The cumulative probability for a z-score of 1.67 is approximately 0.9525.

To find the proportion of people with an IQ between 106 and 125, we subtract the cumulative probability corresponding to the lower z-score from the cumulative probability corresponding to the higher z-score:

0.9525 - 0.6554 = 0.2971

This means that approximately 29.71% of the population falls within the IQ range of 106 to 125.

To find out how many people out of the randomly selected 1450 would have an IQ in this range, we multiply the proportion by the sample size:

0.2971 * 1450 ≈ 431.15

Rounding to the nearest whole number, we find that approximately 431 people would have an IQ between 106 and 125 out of the randomly selected sample of 1450 individuals.