Question

A style of tennis racket comes in a midsize version and an oversize version. Sixty percent of all customers at a certain store want the oversize version. Suppose 10 customers enter the store to buy one of the two versions of the above racket.

a) What is the probability that the number of customers wanting the oversized racket is within 1 standard deviation of the expected number of customers wanting the oversized racket?

b) The store has eight rackets of each version. What is the probability that all 10 customers get the version they want from the store’s current stock?

Answer 1

a)

here this is binomial with parameter n=10 and p=0.6

mean E(x)=μ=np=6

standard deviation σ=√(np(1-p))=1.5492

1 standard deviation from mean values are μ -/+ 1*σ=(4.45,7.55)

therefore probability that the number of customers wanting the oversized racket is within 1 standard deviation of the expected number of customers wanting the oversized racket

=P(5 <=X<=7)=P(X=5)+P(X=6)+P(X=7)

=(₁₀C₅)*(0.6)⁵(.4)⁵+(₁₀C₆)*(0.6)⁶(.4)⁴+(₁₀C₇)*(0.6)⁷(.4)³= 0.2007+0.2508+0.2150 =0.6665~ 0.666

b)

probability that all 10 customers get the version they want from the store’s current stock

P(2<=X<=8)=

∑x=ab     (nCx)px(1−p)(n-x)    =

0.9520