Question

Q4. A nationwide supermarket chain has launched a new loyalty card system. Customers obtain 1 point for each £10 pounds they spend instore or online throughout the year. Points are redeemed as vouchers issued at a rate £1 per 10 points. The number of pounds that a customer spends at this supermarket chain across a year is Normally distributed with a mean of £5000 and standard deviation of £1250. (a) What is the probability that a customer spends more than £5000 in any given year? (b) What percentage of customers receive more than £75 in vouchers each year? (c) What percentage of customers receive between £40 and £75 in vouchers each year? (d) The marketing department want to advertise this loyalty card by saying that "90% of loyalty card holders receive at least £X in vouchers every year". What is the highest number that could be truthfully) used in this statement?

Answer 1

Answer:-

Given that:-

Let,X: Number of pounds spent by the customer at the supermarket chain.

X • N = £5000,0”. £12502)

also, 1point for spent.

是10,

and are redeemed as voucher for 10 points

£1

a) The probability that a customer spends more than in any given year can be calculated as

£5000

by central limit theorem.

P(X > £5000) = P(X-M £5000 - £12500 o

  $P(Z>\frac{\pounds 5000-\pounds 5000}{\pounds 12500})$

$=P(Z>0)$

$P(X>\pounds 5000)=0.5$

b) A customer receives suppose $\pounds k$ in voucher that means customer earned
10 * £k
points that also means he/she spent $10*10*\pounds k$

$\therefore$The percentage of customers receive more than $\pounds 75$ in vouchers each year can be calculated as $P(X>\pounds 7500)*100$

Hence,

P(X £7500)

  $=P(\frac{X-\mu}{\sigma }>\frac{\pounds 7500-\mu}{\sigma })$ by CLT

$=P(Z>\frac{\pounds 7500-\pounds 7500}{\pounds 1250 })$

$=P(Z>2)$

$P(X>\pounds 7500)=0.02275$

$\therefore P(X>\pounds 7500)*100=2.275\%$

$\therefore 2.275\%$ of the customers receive more than $\pounds 75$ in vouchers each year.

c) The percentage of customers receive between $\pounds 40$ and $\pounds 75$ in vouchers each year can be calculated as the same way .

$P(\pounds 4000<X<\pounds 7500)*100$

$\therefore P(\pounds 4000<X<\pounds 7500)$

   by CLT

£4000 - << £7500 - = PG o o

$=P(\frac{\pounds 4000-\pounds 5000}{\pounds 1250 }<Z<\frac{\pounds 7500-\pounds 5000}{\pounds 1250 })$$=P(-0.8<Z<2)$

$\therefore P(\pounds 4000<X<\pounds 7500)=0.76539$$\therefore P(\pounds 4000<X<\pounds 7500)*100=76.539\%$$\therefore76.539\%$ of the customers receive between $\pounds 40$ and $\pounds 75$ in each year.

d) 90% of loyalty card holders receive at least $\pounds$ X in vouchers each year. $\pounds$ X can be caluclated as below.

Let , for the sake or convenience we replace our random variable x as Y.

$P(Y>\pounds X)=0.90$

$\therefore P(\frac{y-\mu}{\sigma }> \frac{\pounds X-\mu}{\sigma })=0.90$ by CLT.

$\therefore P(Z>\frac{\pounds X-\pounds 5000}{\pounds 1250})=0.90$

$\therefore$By using statistic table we get z as z= -1.282

$\therefore \frac{\pounds X-\pounds 5000}{\pounds 1250}=-1.282$

$\therefore \pounds X=\pounds 3397.5$

90% loyalty card holders will receive at least $\pounds 33.975$ every year in voucher.