2020 HSC Mathematics Extension 1
Question 12(b):
Binomial distribution
When a particular biased coin is tossed, the probability of obtaining a head is . This coin is tossed times.
Let be the random variable representing the number of heads obtained. This random variable will have a binomial distribution.
(i) Find the expected value .
(ii) By finding the variance , show that the standard deviation of is approximately .
(iii) By using a normal approximation, find the approximate probability that is between and .
How to recognise this question
Question type: binomial distribution
- The command and notation point to binomial distribution.
- The word “hence” means the later part is intended to reuse the earlier result.
How to handle it: identify the random variable and its parameters before using the matching probability formula.
Watch out: Do not restart the second part with unrelated numbers or discard the result proved in the first part.
Step-by-step answer
- (i) , so .
- NESA (i): 1 mark for the correct expected value.
- (ii) , so .
- NESA (ii): 1 mark for showing via the variance.
- (iii) Using and , the interval is about , i.e. .
- Empirical rule / normal approximation: .
- NESA (iii): 1 mark for the approximate probability (or ).
Common mistake: choosing a familiar formula before checking that its conditions, signs, interval and units match this question.
The paper records 3 total marks for this group. Where the official marking guide provides useful partial-credit criteria, those criteria are stated in the steps above.
Try your hand at it again by answering these questions
Question 1
Let .
(i) Find .
(ii) Show . Hence,
(iii) Using a normal approximation and , estimate as a one-standard-deviation style statement (nearest empirical-rule percentage).
Question 2
Let .
(i) Find .
(ii) By finding , show that the standard deviation of is . Hence,
(iii) By using a normal approximation, find the approximate probability that is between and .
Question 3
Let .
(i) Find .
(ii) By finding , show that the standard deviation of is approximately .
(iii) By using a normal approximation, find the approximate probability that is between and .