21N.1.AHL.TZ0.14

let $X$ be the random variable “the weight of a sack of potatoes”

$\text{P}\left(X<50\right)$                 (M1)

$=0.588 \text{kg} \left(0.587929\dots \right)$                 A1

 

[2 marks]

$\text{P}\left(X<l\right)=0.25$                 (M1)

$49.2 \text{kg} \left(49.1929\dots \right)$                 A1

 

[2 marks]

attempt to sum $10$ independent random variables                 (M1)

$Y=\underset{i=1}{\overset{10}{\text{Σ}}}{X}_i~\text{N}\left(498, 10\times 0.9^2\right)$                 (A1)

$\text{P}\left(Y>500\right)=0.241$                 A1

 

[3 marks]

The first part of the question was often answered well but there were a number of candidates who interpreted finding $P\left(X<50\right)$ by finding $P\left(X<49.9\right)$ or something similar. Not all candidates, however, understood that the lower quartile is given by $P\left(X<l\right)=0.25$. Part (c) was less well understood. Attempts to sum $10$ independent random variables correctly involved multiplication of the mean by $10$ but the standard deviation and not the variance was incorrectly multiplied by $10$.

The first part of the question was often answered well but there were a number of candidates who interpreted finding $P\left(X<50\right)$ by finding $P\left(X<49.9\right)$ or something similar. Not all candidates, however, understood that the lower quartile is given by $P\left(X<l\right)=0.25$. Part (c) was less well understood. Attempts to sum $10$ independent random variables correctly involved multiplication of the mean by $10$ but the standard deviation and not the variance was incorrectly multiplied by $10$.

The first part of the question was often answered well but there were a number of candidates who interpreted finding $P\left(X<50\right)$ by finding $P\left(X<49.9\right)$ or something similar. Not all candidates, however, understood that the lower quartile is given by $P\left(X<l\right)=0.25$. Part (c) was less well understood. Attempts to sum $10$ independent random variables correctly involved multiplication of the mean by $10$ but the standard deviation and not the variance was incorrectly multiplied by $10$.