Question

A particular type of tennis racket comes in a midsize version and an oversize version. Sixty percent f all customers at a certain store want the oversize version. (Round your answers to three decimal places.) (a) Among ten randomly selected customers who want this type of racket, what is the probability that at least five want the oversize version? 0.6331 X (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? 0.6665 (c) 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? 0.8204 X You may need to use the appropriate table in the Appendix of Tables to answer this question. Need Help? Read It Talk to a Tutor

Answer 1

Solution

Let X = number of customers out of the randomly selected 10, who want oversize version. Then,

X ~ B(10, 0.6) [0.6 is 60%] .............................................................................................................................................. (1)

Back-up Theory

If X ~ B(n, p). i.e., X has Binomial Distribution with parameters n and p, where n = number of trials and

p = probability of one success, then, probability mass function (pmf) of X is given by

p(x) = P(X = x) = (ⁿC_x)(p^x)(1 - p)^{n – x}, x = 0, 1, 2, ……. , n …………….....................................................................……..(2)

[This probability can also be directly obtained using Excel Function: Statistical, BINOMDIST…………………………….(2a)

Mean (average) of X = E(X) = µ = np…………………..............................................................……………………………..(3)

Variance of X = V(X) = σ² = np(1 – p)………………………………………….........................................................………..(4)

Standard Deviation of X = SD(X) = σ = √{np(1 – p)} ……………………………………...............................................…...(5)

Now, to work out the solution,

Part (a)

Probability at least 5 want oversize version

= P(X ≥ 5)

= 1 - P(X ≤ 4)

= 1 – 0.1662 [vide (1) and (2a)]

= 0.8338 Answer 1

Part (b)

Vide (3), mean = 10 x 0.6 = 6

Vide (5), standard deviation = √(10 x 0.6 x 0.4) = 1.55

So, within 1 standard deviation of the mean => 4.45 to 6.55 ~ 4 to 7 since X cannot take fractional values.

Then, the required probability

= P(4 ≤ X ≤ 7)

= P(X ≤ 7) – P(X ≤ 4)

= 0.8327 – 0.1662 [vide (1) and (2a)]

= 0.6665 Answer 2

Part (c)

Required probability

= 1 – P(at least one customer does not get the desired version)

The event within brackets above happens when the number of customers who want oversize version is greater than 6 or the number of customers who want midsize version is greater than 6 because up to 6 of each the store can supply. The latter part is equivalent to saying that the number of customers who want oversize version is less than 4.

Then, required probability

= 1 – {P(X > 6) + P(X < 4)}

= 1 – (0.6331 + 0.0548) [vide (1a)]

= 0.3121 Answer 3

DONE