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]