Question

Below are data on 18 people who fell ill from an incident of food poisoning. The data give the incubation period (the time in hours between eating the infected food and the first signs of illness) for each person. 15 16 18 19 20 20 21 28 32 34 36 43 46 46 48 48 72 88 What is the mean incubation period? (Round to 2 decimal places) Find the five number summary to answer the following questions. Do these calculations by hand using the rules given in class. Do not use StatTools. What is the minumum? What is Q1? What is the median? What is Q3? What is the maximum? What is the IQR? Identify any outliers 72 72 and 88 О 88 There are no outliers.

Answer 1

Answer:

$\bar X= \frac{15+16+18+19+20+20+21+28+32+34+36+43+46+46+48+48+72+88}{18}= 36.11$

$Min =15$

$Q_1 = \frac{20+20}{2}=20$

$Median= Q_2= \frac{32+34}{2}=33$

$Q_3 = \frac{46+46}{2}=46$

$Max= 88$

The IQR is $IQR= Q_3 -Q_1= 46-20=26$

We can find the usual limits for the values and we got:

$Lower = Q_1 -1.5 IQR = 20 -1.5*26=-19$

$Upper = Q_3 +1.5 IQR = 46 +1.5*26=85$

So then the potential outliers for this case is just 88>85

Step-by-step explanation:

For this case we have the following data:

15 16 18 19 20 20 21 28 32 34 36 43 46 46 48 48 72 88

The sample size is n =18

We can calculate the mean with the following formula:

$\bar X= \frac{\sum_{i=1}^n X_i}{n}$

And if we replace we got:

$\bar X= \frac{15+16+18+19+20+20+21+28+32+34+36+43+46+46+48+48+72+88}{18}= 36.11$

The minimum for this case is $Min =15$

Now we can find the 5 number summary.

For the first quartile we work with the first 10 observations: 15 16 18 19 20 20 21 28 32 34. And Q1 would be the average between the 5th and 6th position of the data ordered, on this case:

$Q_1 = \frac{20+20}{2}=20$

For the median since we have an even number for the sample size would be the average between the 9th and the 10th position from the dataset ordered and we got:

$Median= Q_2= \frac{32+34}{2}=33$

For the third quartile we work with the last 10 observations: 32 34 36 43 46 46 48 48 72 88. And Q3 would be the average between the 5th and 6th position of the data ordered, on this case:

$Q_3 = \frac{46+46}{2}=46$

The maximum is $Max= 88$

The IQR is $IQR= Q_3 -Q_1= 46-20=26$

We can find the usual limits for the values and we got:

$Lower = Q_1 -1.5 IQR = 20 -1.5*26=-19$

$Upper = Q_3 +1.5 IQR = 46 +1.5*26=85$

So then the potential outliers for this case is just 88>85