Edexcel A level Maths (9MA0) · Statistics › Statistical distributions
Practise The normal 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 1Easy4 marks
The volumes of drink dispensed by a machine into cups, \(V\) ml, are modelled by \(V \sim N(249.4, 2.3^{2})\).
(a) Find \(P(V \gt 252.1)\), giving your answer to 4 decimal places.[2]
(b) Find the proportion of values that are less than 247.8 ml, giving your answer to 4 decimal places.[2]
Show the answer and mark scheme
(a)Answer: 0.1202
M1 for standardising, \(\dfrac{252.1 - 249.4}{2.3}\), or the calculator's normal distribution function with the correct limits
A1 for awrt 0.1202
(b)Answer: 0.2433
M1 for \(P(V \lt 247.8)\)
A1 for awrt 0.2433
Question 2Medium6 marks
The masses of apples from an orchard, \(A\) g, are modelled by \(A \sim N(160, 18^{2})\).
(a) Find \(P(145 \lt A \lt 184)\), giving your answer to 4 decimal places.[2]
(b) In a random sample of 500, find the expected number with values between 145 and 184 g.[1]
(c) Find \(P(A \gt 185 \mid A \gt 145)\).[3]
Show the answer and mark scheme
(a)Answer: 0.7065
M1 for \(P(A \lt 184) - P(A \lt 145)\) or a direct calculator method
A1 for awrt 0.7065
(b)Answer: 353.2
B1ft for \(500 \times\) their probability, awrt 353
(c)Answer: 0.1033
M1 for \(\dfrac{P(A \gt 185)}{P(A \gt 145)}\) (recognising \((A \gt 185) \cap (A \gt 145) = (A \gt 185)\))
A1 for \(P(A \gt 185) = 0.0824\) and \(P(A \gt 145) = 0.7977\)
A1 for awrt 0.1033
Question 3Hard9 marks
The times, \(T\) minutes, that customers spend in a supermarket are modelled by a normal distribution. 10% of customers spend less than 40 minutes in the supermarket, and 20% spend more than 70 minutes.
(a) Find the mean and the standard deviation of \(T\).[5]
(b) Find the probability that a randomly chosen customer spends between 50 and 60 minutes in the supermarket.[2]
(c) A second supermarket models the times its customers spend in the shop by \(\mathrm{N}(12, 8^2)\). Explain why this model is not suitable.[1]
(d) Suggest one feature of real shopping times that a normal model might not capture.[1]
Show the answer and mark scheme
(a)Answer: Mean 58.1 minutes, standard deviation 14.1 minutes
B1 for \(z = -1.2816\) and \(z = 0.8416\) (awrt \(-1.28\) and \(0.84\))
M1 for \(P(50 \lt T \lt 60)\) with their \(\mu\) and \(\sigma\)
A1 for awrt 0.270 or 0.271
Worked solution: \(P(50 \lt T \lt 60) = P(-0.574 \lt Z \lt 0.134) = 0.270\).
(c)Answer: \(P(T \lt 0) = P(Z \lt -1.5) = 0.0668\): the model gives a noticeable probability of a negative time, which is impossible.
B1 for showing that \(P(T \lt 0)\) is not negligible (about 0.067), but times cannot be negative
Worked solution: Under this model \(P(T \lt 0) = P(Z \lt -1.5) = 0.0668\), so about 7% of customers would spend a negative time in the shop, which is impossible.
(d)Answer: Real shopping times are likely to be positively skewed (a few customers stay a very long time), whereas the normal distribution is symmetric.
B1 for skewness (positive skew: a long tail of long visits), or times cannot be negative, or a lower limit
Worked solution: Real shopping times are likely to be positively skewed: most visits are short or moderate, but a few customers stay a very long time. A normal distribution is symmetric, so it cannot capture this.