The weights of the apples are normally distributed with a mean of 200 grams and a standard deviation of 25 grams.

(i)     Given that there are 450 apples on the stall, what is the expected number of apples with a weight of more than 225 grams?

(ii)     Given that 70 % of the apples weigh less than m grams, find the value of m .

The weights of the pears are normally distributed with a mean of ∝ grams and a standard deviation of $\sigma$ grams. Given that 8 % of these pears have a weight of more than 270 grams and 15 % have a weight less than 250 grams, find ∝ and $\sigma$ .

The weights of the plums are normally distributed with a mean of 80 grams and a standard deviation of 4 grams. 5 plums are chosen at random. What is the probability that exactly 3 of them weigh more than 82 grams?

(i)     ${\text{P}}(X > 225) = 0.158...$     (M1)(A1)

expected number $= 450 \times 0.158... = 71.4$     A1

(ii)     ${\text{P}}(X < m) = 0.7$     (M1)

$\Rightarrow m = 213{\text{ (grams)}}$     A1

[5 marks]

$\frac{{270 - \mu }}{\sigma } = 1.40...$     (M1)A1

$\frac{{250 - \mu }}{\sigma } = - 1.03...$     A1 

Note: These could be seen in graphical form.

solving simultaneously     (M1)

$\mu  = 258,{\text{ }}\sigma  = 8.19$     A1A1

[6 marks]

$X \sim {\text{N}}({80,4^2})$

${\text{P}}(X > 82) = 0.3085...$     A1

recognition of the use of binomial distribution.     (M1)

$X \sim {\text{B}}(5,\,0.3085...)$

${\text{P}}(X = 3) = 0.140$     A1

[3 marks]

This was an accessible question for most students with many wholly correct answers seen. In part (b) a few candidates struggled to find the correct values from the calculator and in part (c) a small minority did not see the need to treat it as a binomial distribution.

This was an accessible question for most students with many wholly correct answers seen. In part (b) a few candidates struggled to find the correct values from the calculator and in part (c) a small minority did not see the need to treat it as a binomial distribution.

This was an accessible question for most students with many wholly correct answers seen. In part (b) a few candidates struggled to find the correct values from the calculator and in part (c) a small minority did not see the need to treat it as a binomial distribution.