Farmer Suzie grows turnips and the weights of her turnips are normally distributed with a mean of \(122g\) and standard deviation of \(14.7g\).

(i)     Calculate the percentage of Suzie’s turnips that weigh between \(110g\) and \(130g\).

(ii)     Suzie has \(100\) turnips to take to market. Find the expected number weighing more than \(130g\).

(iii)     Find the probability that at least \(30\) of the \(100g\) turnips weigh more than \(130g\).

Farmer Ray also grows turnips and the weights of his turnips are normally distributed with a mean of \(144g\). Ray only takes to market turnips that weigh more than \(130g\). Over a period of time, Ray finds he has to reject \(1\) in \(15\) turnips due to their being underweight.

(i)     Find the standard deviation of the weights of Ray’s turnips.

(ii)     Ray has \(200\) turnips to take to market. Find the expected number weighing more than \(150g\).

(i)     \(P(110 < X < 130) = 0.49969 \ldots  = 0.500 = 50.0\% \)     (M1)A1

Note:     Accept \(50\)

Note:     Award M1A0 for \(0.50\) (\(0.500\))

(ii)     \(P(X > 130) = (1 - 0.707 \ldots ) = 0.293 \ldots \)     M1

expected number of turnips \( = 29.3\)     A1

Note:     Accept \(29\).

(iii)     no of turnips weighing more than \(130\) is \(Y \sim B(100,{\text{ }}0.293)\)     M1

\(P(Y \ge 30) = 0.478\)     A1

[6 marks]

(i)     \(X \sim N(144,{\text{ }}{\sigma ^2})\)

\(P(X \le 130) = \frac{1}{{15}} = 0.0667\)     (M1)

\(P\left( {Z \le \frac{{130 - 144}}{\sigma }} \right) = 0.0667\)

\(\frac{{14}}{\sigma } = 1.501\)     (A1)

\(\sigma  = 9.33{\text{ g}}\)     A1

(ii)     \(P(X > 150|X > 130) = \frac{{P(X > 150)}}{{P(X > 130)}}\)     M1

\( = \frac{{0.26008 \ldots }}{{1 - 0.06667}} = 0.279\)     A1

expected number of turnips \( = 55.7\)     A1

[6 marks]

Total [12 marks]