Question

i. A sample of 50 Indian households had a mean cash income of $12 per day with a standard deviation of $3 per day. A sample of 70 households had a mean cash income of $2000 per day with a standard deviation of $400 per day. Which sample has the wider spread in cash income?

ii . For a lab class, student’s lab tests was ranked on a scale of 0 to 10.

8, 5, 5, 7, 3, 5, 7, 5, 9, 4, 1, 2, 7, 6, 1

a. Calculate quartile 1 and interpret answer?

b. Calculate 50^th percentile?

c. Calculate interquartile range and interpret your answer?

Answer 1

i) The standard deviation is a number that measures how far data values are from their mean.

The standard deviation provides a numerical measure of the overall amount of variation in a data set and can be used to determine whether a particular data value is close to or far from the mean.

When the standard deviation is a lot larger than zero, the data values are very spread out about the mean.

In this case, for the sample of 70 households the standard deviation $400 per day which is much larger than the standard deviation $3 per day for a sample of 50 households.

Therefore, the spread of cash income is much wider in the sample of 70 households.

ii) Using R software

> x <- c(8,5,5,7,3,5,7,5,9,4,1,2,7,6,1)
> q1 <- quantile(x, 0.25)
> q1
25%
3.5
> q2 <- quantile(x, 0.5)
> q2
50%
5
> q3 <- quantile(x, 0.75)
> q3
75%
7
> i <- q3 - q1
> i
75%
3.5

a. First quartile = Q1 = 3.5

b. 50th percentile = Q2 = Median = 5

c. Inter-quartile range = Q3 - Q1 = 3.5