Question
A farmer grows two types of apples—cooking apples and eating apples. The weights of the apples, in grams, can be modelled as normal distributions with the following parameters.
For each type of apple you can assume that 95% of the weights are within two standard deviations of the mean.
(a) Find the percentage of eating apples that have a weight greater than 140 g.
The farmer grows a large number of apples of which 80% are eating apples.
Both types of apples are picked and randomly mixed together in a cleaning machine.
After cleaning, the machine separates out those that have a weight greater than 140 g into a container.
(b) An apple is randomly selected from this container. Find the probability that it is an eating apple. Give your answer in the form $\frac{c}{d}$, where $c, d \in \mathbb{Z}$.
▶️Answer/Explanation
Solution:-
6. (a) 2.5%
(b) P(weight of cooking apples > 140) = 0.5 (50%) (seen anywhere)
recognition of conditional probability in context
$P(\text{eating apple | weight > 140}) = \frac{P(\text{eating apple and weight of eating apple > 140})}{P(\text{weight of apple > 140})}$
$\frac{P(\text{eating apple $\bigcap$ weight of eating apple > 140})}{[P(\text{eating apple $\bigcap$ weight of eating apple > 140}) + P(\text{cooking apple $\bigcap$ weight of cooking apple > 140})]}$
OR
$\frac{P(\text{eating apple}) P(\text{weight of eating apple > 140})}{[P(\text{eating apple}) P(\text{weight of eating apple >140}) + P(\text{cooking apple}) P(\text{weight of cooking apple | 140})]}$
= $\frac{0.8 \times 0.025}{0.8 \times 0.025 + 0.2 \times 0.5} = \frac{80 \times 2.5}{80 \times 2.5 + 20 \times 50} = \frac{200}{1200} = \frac{1}{6}$
Question
Let X be normally distributed with mean 100 cm and standard deviation 5 cm.
a.On the diagram below, shade the region representing \({\rm{P}}(X > 105)\) .[2]
b.Given that \({\rm{P}}(X < d) = {\rm{P}}(X > 105)\) , find the value of \(d\) . [2]
c.Given that \({\rm{P}}(X > 105) = 0.16\) (correct to two significant figures), find \({\rm{P}}(d < X < 105)\) . [2]
▶️Answer/Explanation
Markscheme
a.
A1A1 N2
Note: Award A1 for vertical line to right of mean, A1 for shading to right of their vertical line.
evidence of recognizing symmetry (M1)
e.g. \(105\) is one standard deviation above the mean so \(d\) is one standard deviation below the mean, shading the corresponding part, \(105 – 100 = 100 – d\)
\(d = 95\) A1 N2
[2 marks]
evidence of using complement (M1)
e.g. \(1 – 0.32\) , \(1 – p\)
\({\rm{P}}(d < X < 105) = 0.68\) A1 N2
[2 marks]
Question
The random variable X is normally distributed with mean 1000 and standard deviation 50.
(a) Find (i) \(P(X < 925)\) (ii) \(P(925 < X < 1025)\) (iii) \(P(X > 1025)\)
(b) Sketch a graph representing the information in \((a)\)
(c) Find the standardised values of \(925\) and \(1025\)
(d) Sketch the corresponding graph of standardised values
(e) Find \(E(X^{2})\)
▶️Answer/Explanation
Ans
(a) (i) \(P(X<925)=0.0668\) (ii) \(P(925<X<1025)\) (iii) \(P(X>1025)=0.309\)
(b) 
(c) \(-1.5\) and \(0.5\) respectively
(d) similar to the above the vertical boundaries are \(-1.5\) and \(0.5\)
(e) \(E(X^{2})=Var(X)+E(X)^{2}=\sigma ^{2}+\mu ^{2}=50^{2}+1000^{2}=1002500\)
Question
The random variable \(X\) is normally distributed with mean \(1000\) and standard deviation \(50\).
(a) 30% is more than \(a\). Find \(a\)
(b) 57% is less than \(b\). Find \(b\)
(c) 60% is between \(c\) and \(d\), where \(c\) and \(d\) are symmetric about the mean. Find \(c\) and \(d\)
(d) Find the interquartile range.
▶️Answer/Explanation
Ans
\(a =1026\) \(b=1008.8\) \(c=958\) and \(d=1042\) IQR=\(1033.7-966.3=67.4\)