Question

Listed below are the durations​ (in hours) of a simple random sample of all flights of a space shuttle program. Find the​ range, variance, and standard deviation for the sample data. Is the lowest duration time​ unusual? Why or why​ not?

76   

90   

234   

195   

165   

265   

197   

372   

252   

237   

384   

340   

225   

245   

0

Answer 1

For the given data the range is calculated as Maximum value - Minimum value

a) Range = 384 - 0 = 384

b) Now to calculate the variance of the dataset we need to find the mean as:

Mean = (0 + 76 + 90 + 165 + 195 + 197 + 225 + 234 + 237 + 245 + 252 + 265 + 340 + 372 + 384)/15
= 3277/15
Mean = 218.4667

b) and the variance as

= (1/15 - 1) x ((0 - 218.4667)2 + ( 76 - 218.4667)2 + ( 90 - 218.4667)2 + ( 165 - 218.4667)2 + ( 195 - 218.4667)2 + ( 197 - 218.4667)2 + ( 225 - 218.4667)2 + ( 234 - 218.4667)2 + ( 237 - 218.4667)2 + ( 245 - 218.4667)2 + ( 252 - 218.4667)2 + ( 265 - 218.4667)2 + ( 340 - 218.4667)2 + ( 372 - 218.4667)2 + ( 384 - 218.4667)2)
= (1/14) x ((-218.4667)2 + (-142.4667)2 + (-128.4667)2 + (-53.4667)2 + (-23.4667)2 + (-21.4667)2 + (6.5333)2 + (15.5333)2 + (18.5333)2 + (26.5333)2 + (33.5333)2 + (46.5333)2 + (121.5333)2 + (153.5333)2 + (165.5333)2)
= (0.0714) x ((47727.69900889) + (20296.76060889) + (16503.69300889) + (2858.68800889) + (550.68600889) + (460.81920889) + (42.68400889) + (241.28340889) + (343.48320889) + (704.01600889) + (1124.48220889) + (2165.34800889) + (14770.34300889) + (23572.47420889) + (27401.27340889))
= (0.0714) x (158763.73333335)
= 11335.73

c) and the sample standard devaiation as:

s =√s2 hence the standard deviation as

= √(11335.730560001)
= 106.4907
d) Since the unusual value is calculated using the Z score which is calculated as:

Where when Z is les than -2 or greater than 2 we say that the value is unusual.

Now as we can see that the Z score is less than 2 hence the lowest value is unusual.'