Porisma

Worked solution

Use a noninteger mean, compare modal counts and audit finite-population assumptions

One question in a survey is as follows:

“Do you use Facebook during office hours at least once a day? Yes or No?”

It is known that on average, 7 out of 9 staff members answered “yes” to the above question.

For a random sample of 60 staff members,

(i) find the standard deviation of the number of staff members who answer “yes”, mark

(ii) find the probability that the number of staff members who answer “yes” exceeds the mean, marks

(iii) find the most probable number of staff members who answer “yes”, marks

(iv) state, with a reason, if the above answers are valid if the total number of staff members in the company is small. mark

[2013/DHS/10(b)(modified)]

CJC · Review Problems · Question 13

Answer

(i) 3.223.22; (ii) 0.5320.532; (iii) 4747; (iv) the source finite-population sampling approximation is not justified for a small workforce.

(i)

Under the source binomial model, X∼B(60,7/9)X\sim B(60,7/9). Then Var⁡(X)=60(7/9)(2/9)=280/27\operatorname{Var}(X)=60(7/9)(2/9)=280/27, so the standard deviation is 280/27≈3.220305944=3.22\sqrt{280/27}\approx3.220305944=3.22 (3 s.f.).

How did this part go?

(ii)

The mean is 60(7/9)=462360(7/9)=46\frac23. Since X is integer-valued, exceeding the mean means X≥47X\geq47. Thus P(X>4623)=1−P(X≤46)≈0.5321291330=0.532P(X>46\frac23)=1-P(X\leq46)\approx0.5321291330=0.532 (3 s.f.). The source complement value is .46787, rounded.

How did this part go?

(iii)

The printed comparison table is:

xxP(X=x)P(X=x)
460.1185
470.1235
480.1171

The most probable number is 47. As an authored modal check, (60+1)(7/9)=4749(60+1)(7/9)=47\frac49 is nonintegral, so the unique binomial mode is 47.

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(iv)

The source answers no: for a small fixed workforce sampled without replacement, the binomial approximation is not justified because successive response indicators are dependent and the conditional probability of “yes” can change materially. The key describes this as the probability not being approximately constant. More precisely, individual marginal probabilities need not change; independence and conditional probabilities are the issue. A fixed finite workforce with a known number of yes responses would instead give a hypergeometric count.

How did this part go?

Common mistakes

A fixed population sampled without replacement creates dependence even when each individual has the same marginal selection probability.

Solution notes

The source finite-population explanation is retained with an explicit marginal-versus-conditional probability qualification.

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