Examples

Question: A year ago, Jennifer began working at a software store. Her supervisor asked her to keep a record of the number of sales she made each month.

The following data set is a list of her sales for the last 12 months:

34, 47, 1, 15, 57, 24, 20, 11, 19, 50, 28, 37

Use Jennifer's sales records to find:

a) the median
b) the range
c) the upper and lower quartiles
d) the interquartile range

Solutions:

a) The values in ascending order are:
1, 11, 15, 19, 20, 24, 28, 34, 37, 47, 50, 57.

Median = (12 + 1) ÷ 2
            = 6.5th value
            = (6^th + 7^th observations) ÷ 2
            = (24 + 28) ÷ 2
            = 26

b) Range = difference between the highest and lowest values
              = 57 - 1
              = 56

c) Lower quartile = value of middle of first half of data Q₁
                           = the median of 1, 11, 15, 19, 20, 24
                           = (3^rd + 4^th observations) ÷ 2
                           = (15 + 19) ÷ 2
                           = 17

Upper quartile = value of middle of second half of data Q₃
                       = the median of 28, 34, 37, 47, 50, 57
                       = (3^rd + 4^th observations) ÷ 2
                       = (37 + 47) ÷ 2
                       = 42

d) Interquartile range = Q₃–Q₁
                                = 42 - 17
                                = 25

Question: Consider a population of voters in a given state. The true proportion of voters who favor candidate A is equal to 0.40. Given a sample of 200 voters, what is the probability that more than half of the voters support candidate A?

Solutions:

Let X denote the count of voters in the sample of 200 who support candidate A. X is distributed as Binomial(200,0.4). The mean of the distribution is equal to 200*0.4 = 80, and the variance is equal to 200*0.4*0.6 = 48. The standard deviation is the square root of the variance, 6.93. The probability that more than half of the voters in the sample support candidate A is equal to the probability that X is greater than 100, which is equal to 1- P(X< 100).

To use the normal approximation to calculate this probability, we should first acknowledge that the normal distribution is continuous and apply the continuity correction. This means that the probability for a single discrete value, such as 100, is extended to the probability of the interval (99.5,100.5). Because we are interested in the probability that X is less than or equal to 100, the normal approximation applies to the upper limit of the interval, 100.5. If we were interested in the probability that X is strictly less than 100, then we would apply the normal approximation to the lower end of the interval, 99.5.

Applying the continuity correction and standardizing the variable X gives the following:

1 - P(X< 100)
= 1 - P(X< 100.5)
= 1 - P(Z< (100.5 - 80)/6.93)
= 1 - P(Z< 20.5/6.93)
= 1 - P(Z< 2.96)
= 1 - (0.9985)
= 0.0015.

Since the value 100 is nearly three standard deviations away from the mean 80, the probability of observing a count this high is extremely small.