All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

Firstly, the number of different friends is a fixed constant.
Secondly, the probability of successfully contacting a friend is constant.

The probability of contacting a friend will vary from friend to friend, as some friend might not be free or contactable during that evening.

\mathrm{R} \sim \mathrm{B}(8,0.7)

\mathrm{P}(R \ge 6) = 1 - \mathrm{P}(R \le 5) \approx 0.552

\mathrm{R} \sim \mathrm{B}(40, 0.7)

Since n = 40 \mathrm{~is~large,~} np = 28 > 5 \mathrm{~and~} nq = 12 > 5 ~ \mathrm{R} \sim \mathrm{N}(28, 8.4) \mathrm{approximately}

\mathrm{P}(R \textless 25)

= \mathrm{P}(R \textless 24.5) by continuity correction

\approx 0.114

KS Comments:

Students should be careful to write all assumptions in the context of question. I cannot stress the importance of this. Conditions for Binomial Approximation have to be well stated and explained. Lastly, remember to do continuity correction.

One Comment

Leave a Reply