As a statistician, you are asked to give advice on designing a volunteer appreci
ID: 3292887 • Letter: A
Question
As a statistician, you are asked to give advice on designing a volunteer appreciation event for a local charity. They plan to put one blue ball and a number of red balls in a black box, let a volunteer draw a ball from the box and award a Starbucks gift card if the volunteer draws the blue ball. They want advice on how many red balls they should put in the box. There will be 400 volunteers attending the event. Their initial plan is to put one blue ball and 9 red balls in the black box.
(a) Under the charity’s initial plan, identify the mean and the standard deviation of the proportion of the volunteers who manage to pick the blue ball among 400 volunteers. [1 mark]
(b) Under the charity’s initial plan, how many gift cards they should prepare in order for the number of prepared gift cards to be enough with at least 97.5% probability to award every volunteer who picks the blue ball? [2 marks]
(c) The charity wants 50 gift cards to be enough with at least 97.5% probability to award every volunteer who picks the blue ball. Find the minimum number of red balls they should put in the box in order to satisfy this criterion. [3 marks]
Explanation / Answer
Solution ± A µ2 [1,n]
Let X = Number of volunteers out of 400 volunteers who pick up a blue ball. Then,
X ~ B(400, p), where p = probability a volunteer would pick up a blue ball.
Since initially, there is 1 blue ball out of 10 balls, p = 1/10 = 0.1.
Part (a)
Mean and standard deviation of B(n, p) are respectively, np and {np(1 - p)} and if Y = proportion of volunteers who pick up a blue ball, then mean and standard deviation of Y are respectively, p and {p(1 - p)/n}
Substituting n = 400 and p = 0.1,
the mean and the standard deviation of the proportion of the volunteers who manage to pick the blue ball among 400 volunteers are 0.1 and 0.015 ANSWER
Part (b)
Gift cards to be prepared = expected number of volunteers who pick the blue ball.
Since we want to ensure at least 97.5% probability, we should have
P(X m) = 0.975, where m = number of gift cards to be prepared.
This probability can be worked out by Binomial which can also be approximated by Normal distribution provided both np and np(1 - p) are greater than 10.
In our case, np = 40 and np(1- p) = 36, both of which are greater than 10. So, Normal approximation would be:
P(X m) = P[Z (m - np)/{np(1 - p)}], where Z ~ N(0. 1).
Now, P[Z (m - np)/{np(1 - p)}] = 0.975 =>
(m - np)/{np(1 - p)} = upper 2.5% point of N(0, 1).
Either using Excel Function (as done here) or by referring to Standard Normal Tables, upper 2.5% point can be found to be 1.96.
So, (m - np)/{np(1 - p)} = (m - 40)/{36) = (m - 40)/6 = 1.96 or m = 40 + (6 x 1.96) = 51.76 = 52 (since number of cards cannot be a fraction).
Thus, in order for the number of prepared gift cards to be enough with at least 97.5% probability to award every volunteer who picks the blue ball, the number of cards should be 52 ANSWER
Part (c)
Here we are given m = 50 and our job is to find the number of red balls to be put in the box to satisfy the given conditions.
Let the number of red balls added be k. Then, the new value of p, say p1 = 1/(10 + k) and we should have, (referring to working of Part (b),
(50 - 400p1)/{400p1(1 – p1)} = 1.96 which on simplification would lead to:
400p1 + 39.2{p1(1 – p1)} = 50.
On solving the above, p1 can be found out and the number of red balls (10 + k)p1 = 1 or
k = (1/p1) – 10.
An easier method would be:
We know from Part (b) that at p = 0.1, m = 52. Since 50 just 2 less than 52, p has to increase slightly. Further, since only number of red balls can be changed, increase the number one by one and substitute p1 in the equation of Part (b) and arrive at the answer.
When number of red balls = 10, p1 = 1/11 and m is 47.68. So,
Adding one more red ball to the box would do the job. ANSWER
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