Chance for having a small stone is 23 and the chance

• Notes
• 24
• 100% (14) 14 out of 14 people found this document helpful

This preview shows page 2 - 6 out of 24 pages.

chance for having a small stone is 2/3 and the chance for a large stone is 1/3, the correct way to determine the effectiveness of the treatment would be as a weighted average: Treatment A= (2/3)(.93) + (1/3)(.73)= .86 Treatment B= (2/3)(.87) +(1/3)(.69)= .81 So treatment A is more effective.
4. The histogram below uses a density scale to represent the ages of students in an art class at the local community center.
a. How should the vertical axis be labeled? b. What is the median age, approximately? c. Is the median age less than, greater than, or above the same as the average age? d. About what percent of students are between ages 15 and 25? Solutions:
5. The histogram below uses a density scale to represent the weights, in ounces, of a sample of Oskie dolls (Oskie is Cal’s beloved bear mascot) sold in stores on Telegraph Ave. a. How should the vertical axis be labeled? b. What value should the vertical axis have at the arrow? b. In which .2 ounce bin does the 75 th percentile lie? Solutions:
3
4 b. 60. (0.2*60 = 12) c. The 75 th percentile is between 16.1 and 16.3 ounces. 6. The number of items produced in a factory during a week is a random variable with mean 50. a. What can be said about the probability that this week’s production is at least 100? Hint: Markov’s inequality. b. If the standard variance of a week’s production is known to equal 5, can we obtain a better bound for part (a)? Solution: Let X denote this week’s production. a. P(X 100) ! [ ! ] !"" = !" !"" = ! ! b. P(X 100) ! ( | ! !" | !" ) !" !" ! = ! !"" . This shows that if the variance is known, using Chebyshev’s inequality may give tighter bounds than Markov’s inequality. 7. Let X be an random variable whose distribution is given by the function f ( x ) = e x x 0 0 otherwise # \$ % % & ' ( ( . From this function we can see that X takes only nonnegative values. X has expectation E(X) = 1 and Var(X) = 1. a. Use Markov’s inequality to find an upper bound for ( 3) P X b. Use Chebyshev’s inequality to find an upper bound for ( 3) P X . c. Fun, calculus challenge: Calculate the actual value of ( 3) P X by integrating f(x) from 0 to infinity. Solution a. Using the Markov inequality P(X ! ) ! ( ! ) ! with t = 3 we get the bound P(X ! ) ! ( ! ) ! = ! ! b. Using Chebyshev inequality ! ( | ! ! ( ! ) | ! ) !"# ! ! ! we have to do a little more work and notice that: ! ( | ! ! | ! ) = ! ( ! ! ! !" ! ! ! ) = ! ( ! ! !" ! ! ) = ! ( ! ! ) Since X takes positive values, we can use Chebshev’s inequality to calculate
5 ! ( ! ! ) = ! ( | ! ! | ! ) = ! ( | ! ! ( ! ) | ! ) !"# ( ! ) ! ! = ! ! c. The actual probability of P(X ! ) is calculated by finding the integral under the f(x) curve: ! ( ! ! ) = ! ! ! !" = ! ! ! ! ! !" = ! ! ! ! ! | ! ! = ! ( ! ! ! ) = ! ! ! ! . !"