Question

Suppose the following tables present the number of specimens that tested positive for Type A and Type B influenza in a country during a flu season. Type A 38 100 186 199 253 380 595 966 1611 2638 3845 4931 5183 5367 5890 Type B 59 95 116 143 156 225 271 366 495 696 851 1060 1140 1101 1202 Send data to Excel (a) Find the mean and median number of Type A cases. Round the answers to at least one decimal place. (b) Find the mean and median number of Type B cases. Round the answers to at least one decimal place. (c) A public health official says that there are more than twice as many cases of Type A influenza than Type B. Do these data support this claim

Answer 1

Answer:

(a) Mean and Median of type A

$\bar x = 2145.47$

$Median = 966$

(b) Mean and Median of type B

$\bar x = 531.73$

$Median = 366$

(c) The claim by the public health worker is true.

Step-by-step explanation:

Given

$Type\ A: 38\ 100\ 186\ 199\ 253\ 380\ 595\ 966\ 1611\ 2638\ 3845\ 4931\ 5183\ 5367\ 5890$

$Type\ B: 59\ 95\ 116\ 143\ 156\ 225\ 271\ 366\ 495\ 696\ 851\ 1060\ 1140\ 1101\ 1202$

$n = 15$

Solving (a): The mean and median of A.

Mean is calculated using:

$\bar x = \frac{\sum x}{n}$

$\bar x = \frac{38 +100 +186 +199+ 253+ 380+ 595+ 966 +1611 +2638 +3845+ 4931+ 5183 +5367 +5890 }{15}$

$\bar x = \frac{32182}{15}$

$\bar x = 2145.47$

The median is calculated using:

$Median = \frac{n+1}{2}th$

$Median = \frac{15+1}{2}th$

$Median = \frac{16}{2}th$

$Median = 8th$

The 8th item is: 966

So:

$Median = 966$

Solving (b): The mean and median of B.

Mean is calculated using:

$\bar x = \frac{\sum x}{n}$

$\bar x = \frac{59 +95 +116+ 143+ 156+ 225+ 271+ 366+ 495 +696+ 851+ 1060+ 1140 +1101+ 1202}{15}$

$\bar x = \frac{7976}{15}$

$\bar x = 531.73$

The median is calculated using:

$Median = \frac{n+1}{2}th$

$Median = \frac{15+1}{2}th$

$Median = \frac{16}{2}th$

$Median = 8th$

The 8th item is: 366

So:

$Median = 366$

(c) The claim by the public health worker is true.

To do this, we simply compare the mean value of both types.

For Type A

$\bar x = 2145.47$

For Type B

$\bar x = 531.73$

The claim is:

$Type\ A > 2 * Type B$

$2145.47 > 2 * 531.73$

$2145.47 > 1063.46$

Since the inequality is true, then the claim is true