Question

ng your paper. 1. (Ch. 1) A sample of 30 high school students took part in a simulated fire drill escape exercise, resulting in the accompanying data on time (in seconds) to complete the escape. 202 376 397 349 380 452 343 394 373 402 403 373 380 364 366 364 317 340 394 301 385 391 391 342 387 398 471 470 337 397 mean= 347.63 (a) Calculate the values of the sample mean and median. =380 (b) By how much could the largest value of 471 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? Q3 34 Q,=364 (c) Find the Qi. Q3, and the inner quartile range. (d) Are there any outliers in the data set? If so, what are they? (e) Are there any extreme outliers? If so, what are they? (f) Draw a box plot representing the above data. if the observations are re-expressed in minutes? (g) What are the values of r and was recorded. ens of a certain type of plastic

Answer 1

a) The formula of mean is given by-

ΙΣ: = 1

Here for these sample mean = 374.63 seconds.

For sample median ($\widetilde{x}$) we have to first arrange the data in ascending order. Here total sample observations is 30.

Therefore, N=30. So the median is (N+1)/2, (30+1)/2 = 15.5 Therefore taking the mean of 15 and 16 position, we get median as 380 secs. Q₂=380 secs.

b) In these data the highest value is 471 secs. So, sample median is not affected when the data is arranged in ascending (or descending) order. Hence, if we increase 471 secs to a large amount, sample median will not be changed. It will be 380 secs.

If 471 secs be decreased then we have to notice about the position. Sample median will remain same(380 secs) if the value 471 secs be decrease up to 380 secs. Rather it will get change.

c) Quartile divides the data into 4 equal parts.First quartile is given by Q₁ = (N+1)/4 th position and third quartile is given by Q₃ = (3N+3)/4 th position.

Here, Q₁ = 347.5 secs and Q₃ = 397 secs Inter quartile range (IQR) = Q₃ - Q₁ = (397 - 347.5) secs = 49.5 secs

d or e) In order to be an outlier, the data value must be: larger than Q₃ by at least 1.5 times the interquartile range (IQR), or smaller than Q₁ by at least 1.5 times the IQR Therefore, 1.5 times IQR = 1.5*49.5 = 74.25. Q₁ - 74.25 = 273.25 secs. Hence, the minimum in the dataset 202 seconds is the outlier. There is no outlier in the upper part as (Q₃ + 74.25) = 471.25 secs. No data is above 471.25 secs.

f) The box plot is given by-

Boxplot of Time (on seconds) Time

g)

So house DAV 1 202 secs =3 min 42 secs | 2301 secs = 5 min 1 secs 3) 317 secs - 5 min 17 secs 4337 secs = 5min 37 secs 5) 340 secs = 5 min 40 secs 6) 342 secs - 5 min 42 secs 7) 343 secs - 5 min 43 secs 8) 349 secs - 5 min 49 secs a) 3.64 secs = 6 min 4 secs 10 364 secs = 6 min 4 secs (11) 366 secs & 6 min 6 secs 12) 373 secs = 6 min 13 secs (3) 373 secs = 6 min 13 secs 14) 376 secs : 6 min 16 secs 15) 380 secs 26 min 20 secs (16) 380 secs = 6 min 20 secs 17) 385 secs = 6 min 25 secs 18) 387 secs = 6 min 27 secs 19) 391 secs - 6 min 31 secs 20) 391 secs = 6 min 31 secs 34) 394 secs - 6 min 34 secs 22) 394 secs - 6 min 34 secs 1397 secs - 6 min 37 secs 397 secs = 6 min 37 secs! 398 secs - 6 min 38 secs 402 secs = 6 min 42 secs (27) 408 seas = 6 min 43 secs 28) 452 secs - 7 min 32 sees. 29) 470 secs - 7 min 50 secs 30) 471 secs = 7 min Sisees En = 187 min 39 sec . Mearen = h Inila - 2 x 187 nun 39 sec = 6 min 15 sec Median = x= 6 min 20 sec 32 se