Question

Listed below are the durations​ (in hours) of a simple random sample of all flights of a space shuttle. Find the​ (a) mean,​ (b) median,​ (c) mode, and​ (d) midrange for the given sample data. Is there a duration time that is very​ unusual? How might that duration time be​ explained? 68 92 231 189 169 267 196 381 257 231 385 330 219 239 0 (a) The mean is _____ hours. ​(Round to one decimal place as​ needed.) (b) The median is _____ hours. ​(Round to one decimal place as​ needed.) Is there a duration time that is very​ unusual? How might that duration time be​ explained? (A) ​No, the flights have usual duration times ranging from 0 to over 375 hours. (B) ​Yes, the time of 0 hours is very unusual. It could represent a flight that was aborted. (C) ​No, there is no flight with an unusual duration time. (D) ​Yes, the time of more than 375 hours is very unusual. It could represent a very long flight.

Answer 1

Mean = 212.93

Median = 231

Mode = 231

Mid range = 192.5

(B) ​Yes, the time of 0 hours is very unusual. It could represent a flight that was aborted.

Explanation:

We are given the following data set:

8, 92, 231, 189, 169, 267, 196, 381, 257, 231, 385, 330, 219, 239, 0

Formula:

a)

`Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}`

`Mean =\displaystyle(3194)/(15) = 212.93`

b)

`Median:\\\text{If n is odd, then}\\\\Median = \displaystyle(n+1)/(2)th ~term \\\\\text{If n is even, then}\\\\Median = \displaystyle((n)/(2)th~term + ((n)/(2)+1)th~term)/(2)`

Sorted data:

0, 8, 92, 169, 189, 196, 219, 231, 231, 239, 257, 267, 330, 381, 385

`\text{Median} = (15 + 1)/(2)^(th)\text{ term} = 8^(th)\text{ term} = 231`

Mode is the most frequent observation in the data.

Mode = 231

It repeated 2 times.

`\text{Mid range} = \frac{\text{Maximum value + Minimum value}}{2}\\\\= (385 + 0)/(2) = 192.5`

0 is an outlier to the given dataset. Thus,

Option (B) ​Yes, the time of 0 hours is very unusual. It could represent a flight that was aborted.