Question

Forty percent of households say they would feel secure if they had $50,000 in savings. you randomly select 8 households and ask them if they would feel secure if they had $50,000 in savings. find the probability that the number that say they would feel secure is (a) exactly five, (b) more than five, and (c) at most five.

Answer 1

Answer:

Let X be the event of feeling secure after saving $50,000,

Given,

The probability of feeling secure after saving $50,000, p = 40 % = 0.4,

So, the probability of not  feeling secure after saving $50,000, q = 1 - p = 0.6,

Since, the binomial distribution formula,

$P(x=r)=^nC_r p^r q^{n-r}$

Where, $^nC_r=\frac{n!}{r!(n-r)!}$

If 8 households choose randomly,

That is, n = 8

(a) the probability of the number that say they would feel secure is exactly 5

$P(X=5)=^8C_5 (0.4)^5 (0.6)^{8-5}$

$=56(0.4)^5 (0.6)^3$

$=0.12386304$

(b) the probability of the number that say they would feel secure is more than five

$P(X>5) = P(X=6)+ P(X=7) + P(X=8)$

$=^8C_6 (0.4)^6 (0.6)^{8-6}+^8C_7 (0.4)^7 (0.6)^{8-7}+^8C_8 (0.4)^8 (0.6)^{8-8}$

$=28(0.4)^6 (0.6)^2 +8(0.4)^7(0.6)+(0.4)^8$

$=0.04980736$

(c) the probability of the number that say they would feel secure is at most five

$P(X\leq 5) = P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) + P(X=5)$

$=^8C_0 (0.4)^0(0.6)^{8-0}+^8C_1(0.4)^1(0.6)^{8-1}+^8C_2 (0.4)^2 (0.6)^{8-2}+8C_3 (0.4)^3 (0.6)^{8-3}+8C_4 (0.4)^4 (0.6)^{8-4}+8C_5(0.4)^5 (0.6)^{8-5}$

$=0.6^8+8(0.4)(0.6)^7+28(0.4)^2(0.6)^6+56(0.4)^3(0.6)^5+70(0.4)^4(0.6)^4+56(0.4)^5(0.6)^3$

$=0.95019264$