Question

Q80

A supermarket has one checkout for customers who wish to purchase 9 items or less. The number of items presented at this checkout by a sample of 19 customers were: [5 Marks] 5 8 7 7 6 6 10 8 9 9 9 6 5 9 8 9 5 6 6 (a) Find the median and quartiles for this set of data. (b) Determine the semi-interquartile range. (c) Calculate the mean and standard deviation for this set of data.

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Answer 1

Given, the data of the number of items presented at this checkout is 5 8 7 7 6 6 10 8 9 9 9 6 5 9 8 9 5 6 6.

The sample size (n)= 19.

First of all, arrange the numbers in ascending order

5 5 5 6 6 6 6 6 7 7 8 8 8 9 9 9 9 9 10.

A) Since n=19 which is odd. So, the middle number in data (at position 10) will be median which is equal to 7.

n/4=4.75 ~ 5 & 3n/4= 14.25~ 14.

Hence 1st quartile(Q1)=6 & 3rd quartile(Q3)=9.

B) Semi-interquartile range is (Q3-Q1)/2 = (9-6)/2=3/2=1.5

C) sum of the given data is 138. So, Mean= 138/19= 7.26

Varaince= (1/n)*sum ((x-mean(x))^2) =2.65

Standard deviation= sqrt(var)= 1.63