Question

A company is in the business of finding addresses of long-lost friends. The company claims to have a 70% success rate. Suppose that you have the names of six friends for whom you have no addresses and decide to use the company to track them.

(b) Find the mean and standard deviation of this probability distribution. What is the expected number of friends for whom addresses will be found? (Round your answers to two decimal places.)

μ = friends

σ = friends

(c) How many names would you have to submit to be 96% sure that at least two addresses will be found? (Enter your answer as a whole number.)
names

Answer 1

b)

Mean = Expected Value = μ_x̅ = n•p = (15)(0.25) = 3.75

n•p•(1 - p) = (15)(0.25)(1 - 0.25) = 2.8125

Standard Deviation = σ_x̅ = √ n•p•(1 - p) = √2.8125 = 1.677

c)

To determine the value of n it is provided that P( X ≥ 2) = 0.96

P( X ≥ 2) = 1 - P( X <2)
= 1 - P( X ≤ 1)
The probability less than or equal to done the excel output is 0.96
⇒1−P(X≤1)=0.96
⇒P(X≤1)=1−0.96
⇒P(X≤1)=0.04
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P( X ≥ 2) = 0.96
1 - P( X ≤ 1) = 0.96
P( X ≤ 1) = 1 - 0.96
P( X ≤ 1) = 0.04
P( X =0) + P( X =1) = 0.04
(1 - p)^n +np( 1 -p)^n-1 = 0.04
0.4^n + 0.6n(0.4)^n-1 =0.04
0.4^n (n +n(0.6/0.4)) = 0.04
n= 7