S4.2, S4.3The normal approximation to the binomial; choosing a distribution
Edexcel A level Maths (9MA0) · Statistics › Statistical distributions
Practise The normal approximation to the binomial; choosing a distribution. 1 exam-style questions plus unlimited generated ones on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.
Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.
Question 1Easy1 mark
A student wants to model the number shown when a fair six-sided die is rolled once.
State whether a binomial, a discrete uniform or a normal distribution is the most suitable model.[1]
Show the answer and mark scheme
Answer: Discrete uniform.
B1 for discrete uniform
Question 2Medium7 marks
In a large town, 30% of adults cycle to work. A random sample of 200 adults is taken, and \(X\) is the number in the sample who cycle to work.
(a) Explain why a normal distribution is a suitable approximation to the distribution of \(X\).[1]
(b) Using a normal approximation, find \(P(X \le 50)\).[3]
(c) A student gives the answer to part (b) as \(P(Y \lt 50) = 0.0614\). Explain the student's error.[1]
(d) Use the binomial distribution to find \(P(X \le 50)\), and comment on the accuracy of the approximation in part (b).[2]
Show the answer and mark scheme
(a)Answer: \(X \sim \mathrm{B}(200, 0.3)\): \(n\) is large and \(p\) is not too close to 0 or 1 (\(np = 60\) and \(n(1 - p) = 140\) are both large).
B1 for \(n\) is large and \(p\) is close enough to 0.5 (or \(np\) and \(n(1 - p)\) both greater than 5)
Worked solution: \(X \sim \mathrm{B}(200, 0.3)\). \(n = 200\) is large and \(p = 0.3\) is not close to 0 or 1 (\(np = 60\), \(n(1 - p) = 140\)), so the distribution is roughly symmetric and bell-shaped.
(c)Answer: No continuity correction: \(X\) is discrete, and the value 50 is represented by the interval from 49.5 to 50.5, so \(P(X \le 50) \approx P(Y \lt 50.5)\).
B1 for no continuity correction: \(X\) is discrete, so \(X \le 50\) corresponds to \(Y \lt 50.5\)
Worked solution: The student has not used a continuity correction. \(X\) takes whole-number values; in the continuous approximation the value 50 covers \(49.5 \le y \lt 50.5\), so \(P(X \le 50)\) should be approximated by \(P(Y \lt 50.5)\).
(d)Answer: 0.0695; the approximation (0.0713) is close, within about 0.002.
B1 for awrt 0.0695
B1 for a comment that the approximation with continuity correction is good (error about 0.002), and better than the uncorrected value 0.0614
Worked solution: \(P(X \le 50) = 0.0695\) from the binomial distribution. The approximation 0.0713 is close (an error of about 0.002), whereas without the continuity correction the error would be about 0.008.
Question 3Hard6 marks
The random variable \(X \sim B(100, 0.5)\). The distribution of \(X\) is approximated by \(Y \sim N(50, 25)\).
(a) Use the approximation to estimate \(P(42 \le X \lt 57)\), giving your answer to 4 decimal places.[4]
(b) Find the exact value of \(P(42 \le X \lt 57)\) and comment on the accuracy of the approximation.[2]
Show the answer and mark scheme
(a)Answer: \(P(41.5 \lt Y \lt 56.5) = 0.8586\)
M1 for either continuity correction, \(41.5\) or \(56.5\)
A1 for both: \(41.5 \lt Y \lt 56.5\)
M1 for standardising or a correct calculator method
A1 for awrt 0.8586
(b)Answer: \(P(X \le 56) - P(X \le 41) = 0.8590\); the approximation differs by 0.0004, so it is very accurate here.
B1 for awrt 0.8590
B1 for a sensible comment on the size of the difference