Probability: Standard Deviation and Pic

(a) Suppose we take a random sample of size 100 from a discrete distribution in this manner: A green die and a red die are thrown simultaneously 100 times and let Xi denote the sum of the spots on the two dice on the ith throw, i = 1, 2,...100. Find the probability that the sample mean number of spots on the two dice is less than 7.5.

n = 100 µ = 7 µ[pic] = 7 σ = 2.41 σ[pic] = 2.41 /[pic]

|X |2 |3 |4 |5 |
|P(X=x) |0.3333 |0.3333 |0.3333 | 1.00 |
|P.X |199.9800 |99.9900 |0.0000 | 299.97 |
|P².X | 119,988.00 | 29,997.00 | - | 149,985.00 |

σ² = 149,985.00 - 299.97² = 60,003

σ = [pic] = 244.9551

P([pic]≥320) = 1 - P ( Z < [pic] - µ[pic] ) σ[pic] = 1 - P ( Z < 320-299.97 ) 244.9551/[pic] = 1 - P ( Z < 20.03 ) 14.1425 = 1 - P ( Z < 1.42) = 1 - 0.9222

= 0.0778 is the probability that at least 320 will attend.

1 - NORMDIST(320,299.97,244.9551/SQRT(300),1) = 0.0783

(c) A circuit contains three resistors wired in series. Each is rated at 6 ohms. Suppose, however, that the true resistance of each one is a normally distributed random variable with a mean of 6 ohms and a standard deviation of 0.3 ohm. What is the probability that the combined resistance will exceed 19 ohms? How "precise" would the manufacturing process have to be to make the probability less than 0.005 that the combined resistance of the circuit would exceed 19 ohms?

n = 3 µ = 6 µ[pic] = 6 σ = 0.3 σ[pic] = 0.3 /[pic]

[pic]= 19 / 3

P ([pic]≥ 6.33) = 1- P ( Z < [pic] - µ[pic] ) σ[pic] = 1 - P (Z < 6.33-6 ) 0.3 /[pic] =