Here are the summary statistics for the monthly payroll for an accounting firm: $\text { lowest salary } = \$ 60,000 \text {, }$ $\text { mean salary } = \$ 140,000 \text {, }$ median = $100,000,range = $240,000, IQR = $120,000, $\text { first quartile } = \$ 70,000$ standard deviation = $80,000.
Suppose that business has been good and the company gives every employee a $\$ 10,000$ raise.Give the new value of each of the summary statistics.
A)Minimum: 60,000; Mean: 140,000; Median: 100,000; Range: 240,000; IQR: 120,000; Q1: 80,000; SD: 80,000
B)Minimum: 70,000; Mean: 150,000; Median: 100,000; Range: 240,000; IQR: 120,000; Q1: 80,000; SD: 80,000
C)Minimum: 70,000; Mean: 140,000; Median: 110,000; Range: 240,000; IQR: 120,000; Q1: 80,000; SD: 80,000
D)Minimum: 70,000; Mean: 150,000; Median: 110,000; Range: 240,000; IQR: 120,000; Q1: 80,000; SD: 80,000
E)Minimum: 60,000; Mean: 150,000; Median: 110,000; Range: 240,000; IQR: 120,000; Q1: 80,000; SD: 80,000

Question

Quizplus · Accepted Answer

- The minimum value cannot be greater than the original minimum, so options A and E are eliminated. 
- The mean will increase by the amount of the raise, which is $50,000 for every employee. Since there are no other changes in the data, the mean will increase from $100,000 to $150,000. Option B has the correct mean, but the minimum value is too high. 
- The median is not affected by the raise since it is simply the middle value of the data set. Therefore, option C is incorrect since the median should remain at $100,000. 
- The range and IQR will remain the same since they are based on the minimum and maximum values and the quartiles, which are not affected by the raise. Therefore, the only correct option is D.

Here Are the Summary Statistics for the Monthly Payroll for an Accounting