A particular type or tennis racket comes in a midsize version and an oversize ve

Question

A particular type or tennis racket comes in a midsize version and an oversize version, sixty percent or an customers at a certain store want the oversize version. (Round your answers to three decimal places.) Among nine randomly selected customers who want this type of racket, what is the probability that at least five want the oversize version? Among ten randomly selected customers, what is the probability that the number who want the oversize version is within 1 standard deviation of the mean value? The store currently has six rackets of each version. What is the probability that all of the next ten customers who want this racket can get the version they want from current stock?

Explanation / Answer

a)

This is a binomial distribution.

n=9 (number of customers)

p = .6 (probability that 'a' customer would want an oversize version)

x = number (out of 9) who wants an oversize version

P( x >= 5)

P( x=k) = 9Ck (.6)^k (.4)^(9 - k)

k=5,6,7,8,9

P(X > =5) = 0.733

b)

P( x>=6) = 0.633
Mean = n*p = 10(.6) = 6

Variance = n*p(1-p) =10(.6)(.4) = 2.4

Standard deviation = sqrt(2.4) = 1.5492

Within 1 standard deviation of the mean is (6 - 1.5492, 6 + 1.1592)
= (4.84, 7.15) = ( 5, 8 )

P( 5 <= x <= 8)
= ( 5 - 6 / 1.5492 < = Z < = 8 - 6 / 1.5492)

= ( - 0.6454 < = Z < = 1.2909)

= 0.6423

c)

P( 4 < = X < = 6)

= P( 4 - 6 / 1.5492 < = Z < = 6 - 6 / 1.5492)

= P( =1.290 <Z < = 0 )

= 0.4016

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