Question

Let \(x\) denote the IQ of an individual selected at random from a certain population. The value of \(x\) must be a whole number. Suppose that the distribution of \(x\) can be approximated by a normal distribution with mean value 100 and standard deviation 15. Approximate the following probabilities. (Hint: See Example 6.32 ) a. \(P(x=100)\) b. \(P(x \leq 110)\) c. \(P(x<110)\) (Hint: \(x<110\) is the same as \(x \leq 109\).) d. \(P(75 \leq x \leq 125)\)

Short answer

a. The probability of \(P(x=100)\) is technically 0 in a continuous distribution. b. The probability of \(P(x \leq 110)\) is 0.7486. c. The probability of \(P(x<110)\) is 0.7257. d. The probability of \(P(75 \leq x \leq 125)\) is 0.9050.

Step-by-step solution

Probability of \(x=100\)

In a continuous probability distribution the probability of any single exact point, including \(x = 100\), is technically 0. This is because there are an infinite number of possibilities for the random variable \(x\), so the probability of any single exact value is infinitesimally small.

Probability of \(x \leq 110\)

The problem stated that \(x\) is approximately normally distributed with a mean of 100 and standard deviation of 15. Therefore, to find the probability of \(x \leq 110\), we first need to convert \(x\) to a z-score: \(z = (x- \mu) / \sigma\), where \(\mu\) is the mean (100 in this case) and \(\sigma\) is the standard deviation (15 in this case). So, \(z = (110-100) / 15 \approx 0.67\). This z-score tells us that 110 is approximately 0.67 standard deviations above the mean. We can lookup this value in a standard normal table or use a calculator function to approximate the probability. Looking up 0.67 on the z-table gives a probability of approximately 0.7486.

Probability of \(x

This is almost similar to the previous step, but since we want the probability of \(x<110\), which is essentially the same as \(x \leq 109\), we need to calculate \(z = (109-100) / 15 \approx 0.60\). Looking up 0.60 on the z-table, the probability is approximately 0.7257.

Probability of \(75 \leq x \leq 125\)

For this part, we need to find the probability of \(x\) being between 75 and 125. We first need to convert these \(x\)-values into z-scores. For \(x = 75\), \(z = (75-100) / 15 \approx -1.67\) and for \(x = 125\), \(z = (125-100) / 15 \approx 1.67\). Looking up these z-scores in the z-table gives us probabilities approximately 0.0475 and 0.9525 respectively. The probability between these z-scores is the difference of these two probabilities. So, \(P(75 \leq x \leq 125) = 0.9525 - 0.0475 \approx 0.9050.