Answer to Question #155725 in Statistics and Probability for shaheen

Question

A sample of 26 offshore oil workers took part in a simulated escape exercise, resulting in the accompanying data on time (sec) to complete the escape: 373 370 364 366 364 325 339 393 356 359 363 375 424 325 394 402 392 369 374 359 356 403 334 397 a. Calculate the values of the sample mean and median. b. By how much could the largest time, currently 424, be increased without affecting the value of the sample median? By how much could this value be decreased without affecting the value of the sample median? c. What are the values of x̄ and the median when the observations are re-expressed in minutes?

Answer 1

a) To find the sample mean we rewrite the given data in ascending order such as:

$325,325,334,339,356,356,359,359,363,364,364,366,369,370,373,374,375,392,393,394,397,402,403,$

$424$

The formula for mean is $\bar x =\frac {\Sigma x_i}{N}$ , where $x_i=$ given sample data , $N=$ number of observations

Here $\Sigma x_i=8876$ , $N=24$

$\therefore \bar x = \frac{8876}{24}=369.8\dot3$

Here number of observations $N=24$ , which is even.

Therefore formula for median is $M=\frac {1}{2}[(\frac{N}{2})$th term $+(\frac {N}{2}+1)$th term $]$

$\therefore M= \frac {1}{2}[12$ th term $+13$ th term $]$

$=\frac {1}{2}[366+369]=\frac {735}{2}=367.5$

b)In this problem, since median is the average of 12th and 13th term. Therefore any increase of the largest value , i.e 424,which is the 24th term , could not effect the value of the sample median.

But, the largest value could be decreased upto 13th term,i.e 369 without effecting the value of sample median.

If it is less than 369 , then 13th term will be change.which will effect the value of sample median.

Therefore largest time could be decreased by $(424-369)$sec,i.e $55$ sec without effecting the value of sample median.

c) When the observations are re-expressed in minutes then new mean $\bar X=\frac{\frac{1}{60} \Sigma x_i}{N}$ $=\frac{\Sigma x_i}{60.N}$

$\therefore \bar X= \frac{8876}{60×24}=\frac{8876}{1440}=6.16$

Again new median $M'=\frac{M}{60}=\frac {367.5}{60} =6.125$