# Probability and Statistics

## The Mendelian theory states that the number of a type of peas that fall into the classifications round and yellow, wrinkled and yellow, round and green, and wrinkled and green should be in the ratio 9:3:3:1. Suppose that 100 such peas revealed 56, 19, 17, and 8 in the respective categories. Are these data consistent with the model? Use α = .05. (The expression 9:3:3:1 means that 9/16 of the peas should be round and yellow, 3/16 should be wrinkled and yellow, etc.)

The Mendelian theory states that the number of a type of peas that fall into the classifications round and yellow, wrinkled and yellow, round and green, and wrinkled and green should be in the ratio 9:3:3:1. Suppose that 100 such peas revealed 56, 19, 17, and 8 in the respective categories. Are these data consistent …

## Your portfolio is invested 30 percent each in A and C, and 40 percent in B. What is the expected return of the portfolio? (Do not round intermediate calculations. Enter your answer as a percent rounded to 2 decimal places, e.g., 32.16.) What is the variance of this portfolio? (Do not round intermediate calculations and round your answer to 5 decimal places, e.g., 32.16161.) What is the standard deviation? (Do not round intermediate calculations. Enter your answer as a percent rounded to 2 decimal places, e.g., 32.16.

EXPERT ANSWER Investment in Stock A =0.30 Investment in Stock A =0.40 Investment in Stock C=0.30 Boom E(Rp)= 0.30(0.34) +0.40( 0.44) +0.30(0.35) = 0.102+ 0.176+ 0.105 =0.3830 or 38.30% Good E(Rp)= 0.30(0.18) +0.40( 0.15) +0.30(0.09) = 0.054+ 0.06+ 0.027 = 0.1410 or 14.10% Poor E(Rp)= 0.30(-0.02) +0.40( -0.05) +0.30(-0.04) = -0.006+ -0.02+ -0.01 = -0.0380 …

## 6. The monthly maintenance cost X of a machine is an exponential random variable with unknown parameter. Studies have determined that P(X 100) 0.64. For a second machine the cost Y is a random variable such that Y-2X. Find P(Y>100) Answer: P(Y 100) 0.8

EXPERT ANSWER 6) let paramter for exponential distribution is  therefore P(X>100)=e-x =e-*100=0.64 hence P(Y>100) =P(2X>100) =P(X>100/2) =e-*100/2 =(e-*100)1/2 =(0.64)1/2 =0.8

## Suppose that n = 100 random samples of water from a freshwater lake and calcium concentration (milligrams per liter) measured. A 95% Cl on the mean calcium concentration is 0.49 <= u <= 0.82.

Suppose that n = 100 random samples of water from a freshwater lake and calcium concentration (milligrams per liter) measured. A 95% Cl on the mean calcium concentration is 0.49 <= u <= 0.82. EXPERT ANSWER

## A confidence interval estimate is desired for the gain in a circuit on a semiconductor device. Assume that gain is normally distributed with standard deviation s = 25. (a) How large must n be if the length of the 95% CI is to be not greater than 50? (b) How large must n be if the length of the 99% CI is to be not greater than 50?

A confidence interval estimate is desired for the gain in a circuit on a semiconductor device. Assume that gain is normally distributed with standard deviation s = 25. (a) How large must n be if the length of the 95% CI is to be not greater than 50? (b) How large must n be if …

## A confidence interval estimate is desired for the gain in a circuit on a semiconductor device. Assume that gain is normally distributed with standard deviation s = 15.

A confidence interval estimate is desired for the gain in a circuit on a semiconductor device. Assume that gain is normally distributed with standard deviation s = 15. (a) How large must n be if the length of the 95% CI is to be not greater than 30? (b) How large must n be if …