| Date | November 2015 | Marks available | 3 | Reference code | 15N.2.sl.TZ0.10 |
| Level | SL only | Paper | 2 | Time zone | TZ0 |
| Command term | Find | Question number | 10 | Adapted from | N/A |
Question
The masses of watermelons grown on a farm are normally distributed with a mean of \(10\) kg.
The watermelons are classified as small, medium or large.
A watermelon is small if its mass is less than \(4\) kg. Five percent of the watermelons are classified as small.
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.
Markscheme
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.
