Question

Exercise \(7.8\) gave the following probability distribution for \(x=\) the number of courses for which a randomly selected student at a certain university is registered: $$ \begin{array}{lrrrrrrr} x & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ p(x) & .02 & .03 & .09 & .25 & .40 & .16 & .05 \end{array} $$ It can be easily verified that \(\mu=4.66\) and \(\sigma=1.20\). a. Because \(\mu-\sigma=3.46\), the \(x\) values 1,2 , and 3 are more than 1 standard deviation below the mean. What is the probability that \(x\) is more than 1 standard deviation below its mean? b. What \(x\) values are more than 2 standard deviations away from the mean value (i.e., either less than \(\mu-2 \sigma\) or greater than \(\mu+2 \sigma)\) ? What is the probability that \(x\) is more than 2 standard deviations away from its mean value?

Short answer

a. The probability that \(x\) is more than 1 standard deviation below its mean is 0.14. b. The \(x\) values that are more than 2 standard deviations away from the mean are 1, 2 and 7. The probability that \(x\) is more than 2 standard deviations away from its mean value is 0.1.

Step-by-step solution

Calculate the probability for part a

The \(x\) values that are more than 1 standard deviation below the mean are 1, 2, and 3. The corresponding probabilities are 0.02, 0.03, and 0.09, respectively. Calculate the total probability by adding these up: \(0.02 + 0.03 + 0.09 = 0.14.\)

Identify the \(x\) values for part b

More than 2 standard deviations from the mean implies values that are less than \( \mu - 2\sigma = 4.66 - 2*1.20 = 2.26 \) or greater than \( \mu + 2\sigma = 4.66 + 2*1.20 = 7.06 \). Therefore, the \(x\) values less than 2.26 are 1 and 2 and \(x\) value greater than 7.06 is 7.

Calculate the probability for part b

The corresponding probabilities for \(x\) values 1, 2 and 7 are 0.02, 0.03, and 0.05, respectively. Calculate the total probability by adding these up: \(0.02 + 0.03 + 0.05 = 0.1.\)