Question

1. iq scores are normally distributed with a mean of 100 and a standard deviation of 15. find the following probabilities. express both probabilities as percentages rounded to two decimal places.

a. the probability that a single randomly chosen person from the population will have an iq score greater than 120 is ________________%.
b. the probability that a sample of 3 individuals will have an average iq score greater than 120 is ____________%

Explanation:

Step1: Calculate the z - score for part a

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(x = 120\), \(\mu=100\), and \(\sigma = 15\).

$$z=\frac{120 - 100}{15}=\frac{20}{15}\approx1.33$$

We want to find \(P(X>120)\), which is \(1 - P(X\leq120)\). Using the standard normal table, \(P(Z\leq1.33)=0.9082\). So \(P(X > 120)=1 - 0.9082=0.0918\).
Converting to a percentage: \(0.0918\times100 = 9.18\%\)

Step2: Calculate the z - score for part b

For a sample of size \(n = 3\), the standard deviation of the sample mean \(\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}\). Here, \(\sigma = 15\) and \(n = 3\), so \(\sigma_{\bar{x}}=\frac{15}{\sqrt{3}}\approx8.66\)
The z - score formula for the sample mean is \(z=\frac{\bar{x}-\mu}{\sigma_{\bar{x}}}\), with \(\bar{x}=120\), \(\mu = 100\)

$$z=\frac{120-100}{\frac{15}{\sqrt{3}}}=\frac{20}{\frac{15}{\sqrt{3}}}\approx2.31$$

We want to find \(P(\bar{X}>120)\), which is \(1 - P(\bar{X}\leq120)\). Using the standard normal table, \(P(Z\leq2.31)=0.9896\). So \(P(\bar{X}>120)=1 - 0.9896 = 0.0104\)
Converting to a percentage: \(0.0104\times100=1.04\%\)

Answer:

a. \(9.18\%\)
b. \(1.04\%\)