Question

1 A small shop has investigated its customers to determine how much money they spend in shop. The study revealed that the spending distribution is approximately normally distributed with a mean of OR 4.11 and standard deviation of OR 1.37.

(i) What percentage of customers will spend less than OR 3.1 in the shop.

(ii) What spending amount corresponds to the top 87th percentile?

(iii) Suppose owner takes a random sample of 25 customers and record their spending. What is the probability that his sample average will spend less than OR 3.1 in the shop?

(iv) Compare the results compared in (i) and (iii) above.

Answer 1

Given = 4.11, = 1.37

To find the probability, we need to find the z scores.

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(a) n = 1 For P( X < 3.1)

Z = (3.1 – 4.11) / [1.37 / sqrt(1)] = 0.74

The required probability from the normal distribution tables is = 0.2296

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(b) The Top 87th Percentile

To find P(X < x) = 0.87

From the standard normal distribution table, the z score at a p value of 0.87 Is 1.13

Therefore 1.13 = (X – 4.11) / 1.37

Solving for X, we get X = (1.13 * 1.37) + 4.11 = 5.66

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(c) n = 1 For P( X < a)

Z = (3.1 – 4.11) / [1.37 / sqrt(1)] = -3.69

The required probability from the normal distribution tables is = 0.0001

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(d) In comparison, we see that the probability has decreased from 0.2296 for 1 customer to 0.0001 for 25 customers.

Since the amount spent is lesser than the mean, we get a negative z score, and as the value of the z score decreases, the probability of the same also decreases. Since the sample size gets multiplied in the numerator, the negative value gets further decreased as sample size increases, and hence the probability also decreases.

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