The shopkeeper places $100$ randomly chosen potatoes on a weighing machine. Find the probability that their total weight exceeds $10$ kilograms.

Find the minimum number of randomly selected potatoes which are needed to ensure that their total weight exceeds $10$ kilograms with probability greater than $0.95$.

let $T$ denote the total weight, then

$T \sim N(9800,25600)$    
(M1)(A1)

${\rm{P}}(T > 10000) = 0.106$     A1

[3 marks]

let there be $n$ potatoes, in this case,

$T \sim {\rm{N}}(98n,256n)$     A1

we require

${\rm{P}}(T > 10000) > 0.95$     (M1)

or equivalently

${\rm{P}}(T \le 10000) < 0.05$     A1

standardizing,

${\rm{P}}\left( {Z \le \frac{{10000 - 98n}}{{16\sqrt n }}} \right) < 0.05$     A1

$\frac{{10000 - 98n}}{{16\sqrt n }} < - 1.6449 \ldots$     (A1)

$98n - 26.32\sqrt n  - 10000 > 0$     A1

solving the corresponding equation, $n = 104.7 \ldots$     (A1)

the required minimum value is $105$     A1

Note: Part (b) could also be solved using SOLVER and normalcdf, or by trial and improvement.

Note: Allow the use of $=$ instead of $<$ and $>$ throughout.

[8 marks]