Find the standard deviation of the masses of the watermelons.

The following table shows the percentages of small, medium and large watermelons grown on the farm.

A watermelon is large if its mass is greater than \(w\) kg.

Find the value of \(w\).

All the medium and large watermelons are delivered to a grocer.

The grocer selects a watermelon at random from this delivery. Find the probability that it is medium.

All the medium and large watermelons are delivered to a grocer.

The grocer sells all the medium watermelons for $1.75 each, and all the large watermelons for $3.00 each. His costs on this delivery are $300, and his total profit is $150. Find the number of watermelons in the delivery.

finding standardized value for 4 kg (seen anywhere)     (A1)

eg\(\;\;\;z =  - 1.64485\)

attempt to standardize     (M1)

eg\(\;\;\;\sigma  = \frac{{x - \mu }}{z},{\text{ }}\frac{{4 - 10}}{\sigma }\)

correct substitution     (A1)

eg\(\;\;\; - 1.64 = \frac{{4 - 10}}{\sigma },{\text{ }}\frac{{4 - 10}}{{ - 1.64}}\)

\(\sigma  = 3.64774\)

\(\sigma  = 3.65\)     A1     N2

[4 marks]

valid approach     (M1)

eg\(\;\;\;1 - p,{\text{ 0.62, }}\frac{{w - 10}}{{3.65}} = 0.305\)

\(w = 11.1143\)

\(w = 11.1\)     A1     N2

[2 marks]

attempt to restrict melon population     (M1)

eg\(\;\;\;\)\(95\% \) are delivered, \({\text{P}}({\text{medium}}|{\text{delivered}}),{\text{ }}57 + 38\)

correct probability for medium watermelons     (A1)

eg\(\;\;\;\frac{{0.57}}{{0.95}}\)

\(\frac{{57}}{{95}},{\text{ }}0.6,{\text{ }}60\% \)     A1     N3

[3 marks]

proportion of large watermelons (seen anywhere)     (A1)

eg\(\;\;\;{\text{P(large)}} = 0.4,{\text{ }}40\% \)

correct approach to find total sales (seen anywhere)     (A1)

eg\(\;\;\;150 = {\text{sales}} - 300,{\text{ total sales}} = \$ 450\)

correct expression     (A1)

eg\(\;\;\;1.75(0.6x) + 3(0.4x),{\text{ }}1.75(0.6) + 3(0.4)\)

evidence of correct working     (A1)

eg\(\;\;\;1.75(0.6x) + 3(0.4x) = 450,{\text{ }}2.25x = 450\)

200 watermelons in the delivery     A1     N2

Notes:     If candidate answers 0.57 in part (c), the FT values are \({\text{P(large)}} = 0.43\) and 197 watermelons. Award FT marks if working shown.

Award N0 for 197.