Question

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. (round your answers to three decimal places.) (a) among nine 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 1

This is a binomial probability question with success being that a customer wants the oversize version and the probability of success (p) = 60% = 0.6 and the probability of failure (q) = 1 - p = 1 - 0.6 = 0.4

For a binomial probability,

$P(x)=\, ^nC_xp^xq^{n-x}$

Part A:

n = 9, x ≥ 6

$P(x\geq6)=P(6)+P(7)+P(8)+P(9) \\ \\ =\, ^9C_6(0.6)^6(0.4)^{9-6}+\, ^9C_7(0.6)^7(0.4)^{9-7}+\, ^9C_8(0.6)^8(0.4)^{9-8} \\ +\, ^9C_9(0.6)^9(0.4)^{9-9} \\ \\ =84(0.046656)(0.064)+36(0.0279936)(0.16)+9(0.01679616)(0.4) \\ +1(0.010077696)(1) \\ \\ =0.2508+0.1612+0.0605+0.0101=0.4826$

Therefore, the probability that at least six among nine randomly selected customers who want this type of racket want the oversize version is 0.4826 = 48.3%



Part B:

n = 10, p = 0.6, q = 0.4

mean of a binomial distribution is given by:

$\mu=np=10(0.6)=6$

and the standard deviation of a binomial distribution is given by

$\sigma= \sqrt{npq} = \sqrt{10(0.6)(.04)} = \sqrt{2.4} =1.55$

Thus, the numbers within 1 standard deviation of the mean is given by:

$6-1.55\leq x\leq6+1.55 \\ \\ 4.45 \leq x \leq 7.55$

$P(4.45 \leq x \leq 7.55)=P(5)+P(6)+P(7) \\ \\ =\,^9C_5(0.6)^5(0.4)^{9-5}+\,^9C_6(0.6)^6(0.4)^{9-6}+\, ^9C_7(0.6)^7(0.4)^{9-7} \\ \\ =126(0.07776)(0.0256)+84(0.046656)(0.064)+36(0.0279936)(0.16) \\ \\ =0.2508+0.2508+0.1612=0.6628$

Therefore, the probability that the number who want the oversize version among ten randomly selected customers is within 1 standard deviation of the mean value is 0.6628 = 66.3%.



Part C:

Given that the store currently has seven rackets of each version the probability that all of the next ten customers who want this racket can get the version they want from current stock is the same as the probability that among the next ten customers, between 3 to 7 (inclusive) customers will demand for the oversize version.

n = 10, p = 0.6, q = 0.4

$P(3\leq x\leq7)=P(3)+P(4)+P(5)+P(6)+P(7) \\ \\ =\,^9C_3(0.6)^3(0.4)^{9-3}+^9C_4(0.6)^4(0.4)^{9-4}+^9C_5(0.6)^5(0.4)^{9-5}\\+\,^9C_6(0.6)^6(0.4)^{9-6}+\, ^9C_7(0.6)^7(0.4)^{9-7} \\ \\ =84(0.216)(0.004096)+126(0.1296)(0.01024)+126(0.07776)(0.0256)\\+84(0.046656)(0.064)+36(0.0279936)(0.16) \\ \\ =0.0743+0.1672+0.2508+0.2508+0.1612=0.9043$

Therefore, the probability that all of the next ten customers who want this racket can get the version they want from current stock, given that the store currently has seven rackets of each version is 0.9043 = 90.4%