how do i do this?! thanks! One source reports that 30% of US households live wit

ID: 3359676 • Letter: H

Question

how do i do this?! thanks!

One source reports that 30% of US households live with at least one cat. If four US households are selected at random:

a) Find the probability distribution of the number of US households owning at least one cat.

b) WhatistheprobabilitythatofthefourUShouseholdsselected,thenumberlivingwithoneormorepets

is:

exactly 3?

at least 3?

at most 3?

c) Find and interpret the expected number of the four selected US households owning at least one cat.

here are the answers but how!

b) i. 0.0756, ii. 0.0837, iii. 0.9919
c) We expect 1.2 US households of the 4 to own at least one cat.

Here probability of US households to live with at least one cat.

Sample size n = 4 and p = 0.3 so distribution is a binomial distribution. Where,

(a) If X is the number of US households out of these 4 family sample.

Pr(X) = 4CX(0.30)X (0.7)4-X = BIN(X; 4; 0.3)

(b) Pr (X = 3) = 4C3(0.30)3 (0.7)1 = 0.108

Pr(X >= 3) = Pr(X = 3) + Pr( X = 4) Pr( X = 3,4) = 4C3(0.30)3 (0.7)1 + 4C4(0.30)4 (0.7)0 = 0.1161

Pr(X <= 3) = Pr( X= 0,1,2,3) = 1 - Pr(X = 4) = 1 - 4C4(0.30)4 (0.7)0 = 0.9919

E(X) = np = 4 * 0.3 = 1.2

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