Find the probability that a randomly chosen orange weighs more than 200 grams.

Five of these oranges are selected at random to be put into a bag. Find the probability that the combined weight of the five oranges is less than 1 kilogram.

The farm also produces lemons whose weights may be assumed to be normally distributed with mean 75 grams and standard deviation 3 grams. Find the probability that the weight of a randomly chosen orange is more than three times the weight of a randomly chosen lemon.

$z = \frac{{200 - 205}}{{10}} = - 0.5$     (M1)

probability = 0.691 (accept 0.692)     A1

Note: Award M1A0 for 0.309 or 0.308

[2 marks]

let X be the total weight of the 5 oranges

then ${\text{E}}(X) = 5 \times 205 = 1025$     (A1)

${\text{Var}}(X) = 5 \times 100 = 500$     (M1)(A1)

${\text{P}}(X < 1000) = 0.132$     A1

[4 marks]

let Y = B – 3C where B is the weight of a random orange and C the weight of a random lemon     (M1)

${\text{E}}(Y) = 205 - 3 \times 75 = - 20$     (A1)

${\text{Var}}(Y) = 100 + 9 \times 9 = 181$     (M1)(A1)

${\text{P}}(Y > 0) = 0.0686$     A1

[5 marks]

Note: Award A1 for 0.0681 obtained from tables

As might be expected, (a) was well answered by many candidates, although those who gave 0.6915 straight from tables were given an arithmetic penalty. Parts (b) and (c), however, were not so well answered with errors in calculating the variances being the most common source of incorrect solutions. In particular, some candidates are still uncertain about the difference between nX and $\sum\limits_{i = 1}^n {{X_i}}$ .

As might be expected, (a) was well answered by many candidates, although those who gave 0.6915 straight from tables were given an arithmetic penalty. Parts (b) and (c), however, were not so well answered with errors in calculating the variances being the most common source of incorrect solutions. In particular, some candidates are still uncertain about the difference between nX and $\sum\limits_{i = 1}^n {{X_i}}$ .

As might be expected, (a) was well answered by many candidates, although those who gave 0.6915 straight from tables were given an arithmetic penalty. Parts (b) and (c), however, were not so well answered with errors in calculating the variances being the most common source of incorrect solutions. In particular, some candidates are still uncertain about the difference between nX and $\sum\limits_{i = 1}^n {{X_i}}$ .