Chapter 3.1, Problem 45E

Flu season: The following tables present the number of specimens that tested positive for Type A and Type B influenza in the United States during the first 15 weeks of a recent flu season.

1. Find the mean and median number of Type A cases in the first 15 weeks of the flu season.
2. Find the mean and median number of Type B cases in the first 15 weeks of the flu season.
3. A public health official says that there are more than twice as many cases of Type A influenza as Type B. Do these data support this claim?

To determine

To Find: the mean and median number of type A cases.

Answer to Problem 45E

Number of type A cases in first $15$ weeks of flu season, Mean number $=2145$ and Median number $=999$

Explanation of Solution

Given:

  $36,99,177,200,258,384,584,999,1539,2748,3764,4841,5346,5405,5795$

Formula used:

  $\begin{array}{l}\text{Mean}(\overline{x})=\frac{{\displaystyle \sum x_i}}{n}\\ \\ \text{Median}=\left\{\begin{array}{l}\frac{{\left(n+1\right)}^{\text{th }}\text{term}}{2},\text{When}\text{n}\text{is}\text{odd}\\ \frac{{\left(\frac{n}{2}+1\right)}^{\text{th }}\text{term + }{\left(\frac{n}{2}\right)}^{\text{th}}\text{term}}{2},\text{When n is even}\end{array}\right\}\end{array}$

Calculation:

Mean and Median number of type A cases in first $15$ weeks of flu season, Mean, $\overline{x}$

  $\begin{array}{l}=\frac{x_1+x_2+------x_n}{n}\\ =\frac{36+99+177+200+258+384+584+999+1539+2748+3764+4841+5346+5405+5795}{15}\\ =\frac{32175}{15}=2145\end{array}$

For Median, first we have to arrange data in sequential order, $36,99,177,200,258,384,584,999,1539,2748,3764,4841,5346,5405,5795$

  $\begin{array}{c}\text{Median =}{\left(\frac{\left(\text{n+1}\right)}{\text{2}}\right)}^{\text{th}}\text{term}\\ \text{=}{\left(\frac{\text{15+1}}{\text{2}}\right)}^{\text{th}}\text{term}\\ \text{=}{\left(\frac{\text{16}}{\text{2}}\right)}^{\text{th}}\text{term}\\ {\text{=8}}^{\text{th}}\text{term}\\ \text{=999}\end{array}$

Thus, type A cases in first $15$ weeks of flu season, mean number $=2145$ and median number $=999$ .

To determine

To Find: To Find the mean and median number of type B cases.

Answer to Problem 45E

Number of type B cases in first $15$ weeks of flu season, Mean number $\approx 528.47$ and Median number $=359$

Explanation of Solution

Calculation- Mean and Median number of type B cases in first $15$ weeks of flu season, Mean, $\overline{x}$

  $\begin{array}{l}=\frac{x_1+x_2+------x_n}{n}\\ =\frac{56+98+121+139+151+230+263+359+495+674+821+1096+1162+1106+1156}{15}\\ =\frac{7927}{15}=528.4666\approx 528.47\end{array}$

For Median, arrange data in sequential order, we get, $56,98,121,139,151,230,263,359,495,674,821,1096,1106,1156,1162$

for odd numbers

  $\begin{array}{c}\text{Median =}{\left(\frac{\left(\text{n+1}\right)}{\text{2}}\right)}^{\text{th}}\text{term}\\ \text{=}{\left(\frac{\text{15+1}}{\text{2}}\right)}^{\text{th}}\text{term}\\ \text{=}{\left(\frac{\text{16}}{\text{2}}\right)}^{\text{th}}\text{term}\\ {\text{=8}}^{\text{th}}\text{term}\\ \text{=359}\end{array}$

Thus, type B cases in first $15$ weeks of flu season, mean number $\approx 528.47$ and median number

  $=359$ .

To determine

To Find: whether there are more than twice as many cases as type A influenza as type B.

Answer to Problem 45E

There are more than twice as many cases of type A influenza as type B.

Explanation of Solution

By looking at the mean median number of type A and type B cases, we get,

Number of type A cases in first $15$ weeks of flu season, Mean number $=2145$ and Median number $=999$

Number of type B cases in first $15$ weeks of flu season, Mean number $\approx 528.47$ and Median number $=359$

So, it can be said that there are more than twice as many cases of type A influenza as type B as claimed by a public health official.