A particular type 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. a. ?Among ten randomly selected customers who want this type of racket, what is the probability that at least six want the oversize version? b. ?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? c. ?The store currently has seven 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?

Answer : Step 1 : Given a particular type 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. Let x be the number of people who want an oversize version of a tennis racket. a). Among ten randomly selected customers who want this type of racket. Now we have to calculate the probability that at least six want the oversize version. X ~ Binomial (10,0.60) The formula of the binomial distribution is n P(X) = ( )(p) (1 p) nx x=0 x 10 10 x 10x P(X 6) = ( )(0.x0) (1 0.60) x=6 P(X 6) = ( 60)(0.60) (1 0.60) 156+ ( 70)(0.60) (1 0.60) 157+ ( 80)(0.60) (1 0.60) 158+ 10 9 159 10 10 1510 +( 9 )(0.60) (1 0.60) + ( 10)(0.60) (1 0.60) P(X 6) = 0.2508+0.2150+0.1209+0.0403+0.0060 P(X 6) = 0.6330 Therefore the probability that at least six want the oversize version is 0.6330 Step 2 : b). Among ten randomly selected customers. Now we have to calculate the number who want the oversize version is within 1 standard deviation of the mean value. First find the mean and the standard deviation. Now we are finding the mean. The formula of the mean is E(X) = E(X) = np E(X) = 10 × 0.60 E(X) = 6 Then we have to find variance. The formula of the variance is V (X) = 2 V (X) = np(1 p) V (X) = 6(1 0.60) V (X) = 6(0.40) V (X) = 2.4 Then the formula of the standard deviation is square root of the variance. StdDev(x) = = p(1 p) We know that np(1 p) = 2.4 = 2.4 = 1.5492 Now we have to calculate the probability that the number who want the oversize version is within 1 standard deviation of the mean value. P ( X < ) = P( < X < + ) | | We know that Mean = 6and standard deviation = 1.5492 P ( | < |) = P( 6 1.5492 < X < 6 + 1.5492) P ( | < |) = P( 4.4508 < X < 7.5492) = P( 5 X 7) = P(X = 5) + P(X = 6) + P(X = 7) = 0.2007 + 0.2508 + 0.2150) = 0.6665 Therefore the number who want the oversize version is within 1 standard deviation of the mean value is 0.6665.