Pre-U Pre-U 9794/3 2013 June — Question 4 10 marks

Exam BoardPre-U
ModulePre-U 9794/3 (Pre-U Mathematics Paper 3)
Year2013
SessionJune
Marks10
TopicNormal Distribution
TypeDirect binomial from normal probability
DifficultyStandard +0.3 This is a straightforward application of normal distribution followed by binomial probability. Part (i) is a standard z-score calculation, part (ii) applies binomial distribution using the probability from (i), and part (iii) uses expectation. All steps are routine with no novel insight required, making it slightly easier than average but still requiring multiple techniques.
Spec2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation

4 A tomato grower grows just one variety of tomatoes. The weights of these tomatoes are found to be normally distributed with a mean of 85.1 grams and a standard deviation of 3.4 grams.
  1. Find the probability that a randomly chosen tomato of this variety weighs less than 80 grams.
  2. The grower puts the tomatoes in packs of 6 . Find the probability that, in a randomly chosen pack of 6 , at most one tomato weighs less than 80 grams.
  3. The grower supplies consignments of 250 packs of these tomatoes to a retailer. For a randomly chosen consignment, find the expected number of packs having more than one tomato weighing less than 80 grams.

Question 4
\(X \sim N(85.1, 3.4^2)\)
(i)
\(P\left(Z < \frac{80 - 85.1}{3.4}\right)\) — M1 (Standardising)
\(= \Phi(-1.5) = 1 - \Phi(1.5) = 1 - 0.9332\) — M1 (\(1 - \ldots\) to deal with negative \(z\) value)
\(= 0.0668\) — A1 [3]
(ii) \(P(B(6,\ 0.0668) \leqslant 1)\) — M1 (Recognise need for \(B(6, p)\). Possibly implied by partially correct terms in the next line)
\(= 0.9332^6 + 6 \times 0.9332^5 \times 0.0668\) — M1, M1 (Either term correct; Sum of two correct terms)
\(= 0.66046... + 0.28366...\)
\(= 0.944(12\ ...)\) — A1 [4] (Ft *their* \(p\) from (i))
(iii) \(250 \times (1 - 0.9441)\) — M1, M1 (\(250 \times \ldots\); \((1 - \text{(ii)})\))
\(= 13.975 \approx 14.0\) — A1 [3]
Must be at least 1 dp. Do not allow answer rounded to the nearest integer, even following an answer to 3sf or better.
Total: [10]
**Question 4**

$X \sim N(85.1, 3.4^2)$

**(i)**
$P\left(Z < \frac{80 - 85.1}{3.4}\right)$ — M1 (Standardising)

$= \Phi(-1.5) = 1 - \Phi(1.5) = 1 - 0.9332$ — M1 ($1 - \ldots$ to deal with negative $z$ value)

$= 0.0668$ — A1 **[3]**

**(ii)** $P(B(6,\ 0.0668) \leqslant 1)$ — M1 (Recognise need for $B(6, p)$. Possibly implied by partially correct terms in the next line)

$= 0.9332^6 + 6 \times 0.9332^5 \times 0.0668$ — M1, M1 (Either term correct; Sum of two correct terms)

$= 0.66046... + 0.28366...$

$= 0.944(12\ ...)$ — A1 **[4]** (Ft *their* $p$ from (i))

**(iii)** $250 \times (1 - 0.9441)$ — M1, M1 ($250 \times \ldots$; $(1 - \text{(ii)})$)

$= 13.975 \approx 14.0$ — A1 **[3]**

Must be at least 1 dp. Do not allow answer rounded to the nearest integer, even following an answer to 3sf or better.

**Total: [10]**
4 A tomato grower grows just one variety of tomatoes. The weights of these tomatoes are found to be normally distributed with a mean of 85.1 grams and a standard deviation of 3.4 grams.\\
(i) Find the probability that a randomly chosen tomato of this variety weighs less than 80 grams.\\
(ii) The grower puts the tomatoes in packs of 6 . Find the probability that, in a randomly chosen pack of 6 , at most one tomato weighs less than 80 grams.\\
(iii) The grower supplies consignments of 250 packs of these tomatoes to a retailer. For a randomly chosen consignment, find the expected number of packs having more than one tomato weighing less than 80 grams.

\hfill \mbox{\textit{Pre-U Pre-U 9794/3 2013 Q4 [10]}}