{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

HWKsolutions4

# HWKsolutions4 - STAT35AHW4Solutions...

This preview shows pages 1–2. Sign up to view the full content.

1 STAT 35A HW4 Solutions http://www.stat.ucla.edu/~dinov/courses_students.dir/08/Spring/STAT35.dir Problem 1 a. E(X), the expected value of the R.V. X. E(X) = 0.08 × 0 + 0.15 × 1 + 0.45 × 2 + 0.27 × 3 + 0.053 × 4 = 2.06 b. V(X), the variance of X. V(X) = 0.08 × (0 2.06) 2 + 0.15 × (1 2.06) 2 + 0.45 × (2 2.06) 2 + 0.27 × (3 2.06) 2 + 0.053 × (4 2.06) 2 = 0.9364 c. SD(X), the standard deviation of X. SD(X) = 0.9364 ½ = 0.9677 d. V(X), using the shortcut formula, V(X) = E(X 2 ) μ 2 . Should equal the answer in part (b). V(X) = 0.08 × 0 2 + 0.15 × 1 2 + 0.45 × 2 2 + 0.27 × 3 2 + 0.053 × 4 2 – 2.06 2 = 0.9364 Problem 2 a. Compute the first three moments of the random variable X, E(X), E(X 2 ), E(X 3 ), and V(X). E(X) = 0.2 × 13.5 + 0.5 × 15.9 + 0.3 × 19.1 = 16.38 E(X 2 ) = 0.2 × 13.5 2 + 0.5 × 15.9 2 + 0.3 × 19.1 2 = 272.298 E(X 3 ) = 0.2 × 13.5 3 + 0.5 × 15.9 3 + 0.3 × 19.1 3 = 4592.276 V(X) = 272.298 – 16.38 2 = 3.9936 b. If the price of a freezer having capacity X ft3 is 25X 8.5, what is the expected price paid by the next customer to buy a freezer? E(25X 8.5) = 25 × 16.38 – 8.5 = 401 c. What is the variance of the price 25X 8.5 paid by the next customer? Interpret this value!

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}