Question

A budgeting Web site reported that 15​% of U.S. households have withdrawn money from a​ 401(k) or other retirement account for needs other than retirement in 2013. A random sample of

11 U.S. households was selected. Complete parts a through e below.

a. What is the probability that exactly two households withdrew funds from a retirement account for needs other than retirement? - .2866

b. What is the probability that less than three households withdrew funds from a retirement account for needs other than retirement?

c. What is the probability that more than five households withdrew funds from a retirement account for needs other than retirement?

d. What is the mean for this distribution?

e. What is the standard deviation of this distribution?

Answer 1

Solution :

Given that n = 11, p = 0.15

=> q = 1 - p = 0.85

=> for binomial distribution, P(x = r) = nCr*p^r*q^(n-r)

a.
=> P(x = 2) = 11C2*0.15^2*0.85^9

= 0.2866

b.
=> P(x < 3) = P(x = 0) + P(x = 1) + P(x = 2)

= 11C0*0.15^0*0.85^11 + 11C1*0.15^1*0.85^10 + 11C2*0.15^2*0.85^9

= 0.1673 + 0.3248 + 0.2866  

= 0.7788  

c.
=> P(x > 5) = 1 - P(x <= 5)

= 1 - [P(x = 0) + P(x = 1) + P(x = 2) + P(x = 3) + P(x = 4) + P(x = 5)]

= 1 - [11C0*0.15^0*0.85^11 + 11C1*0.15^1*0.85^10 + 11C2*0.15^2*0.85^9 + 11C3*0.15^3*0.85^8 + 11C4*0.15^4*0.85^7 + 11C5*0.15^5*0.85^6]

= 1 - [0.1673 + 0.3248 + 0.2866 + 0.1517 + 0.0536 + 0.0132]

= 1 - 0.9973  

= 0.0027

d.
=> Mean = n*p

= 11*0.15

= 1.65

e.
=> standard deviation = sqrt(n*p*q)

= sqrt(11*0.15*0.85)

= 1.1843