Question

From 1984 to 1995 , the winning scores for a golf turnament were 276,279,279,277,278,278,280,282,285,272,279 and 278. Using the standart deviation from this sample find the percent of winning scores that fall within one standart deviation of the mean.

Answer 1

Mean = 276 + 279 + 279 + 277 + 278 + 278 + 280 + 282 + 285 + 272 + 279 + 278 = 3343/12 = 278.58 = 279(approx)

To find standard deviation, first we will find variance.

Find the difference between mean and sample and then square it.

276 - 279 = (-3)² = 9

279 - 279 = 0

279 - 279 = 0

277 - 279 = (-2)² = 4

278 - 279 = (-1)² = 1

278 - 279 = (-1)² = 1

280 - 279 = (1)² = 1

282 - 279 = (3)² = 9

285 - 279 = (6)² = 36

272 - 279 = (-7)² = 49

279 - 279 = 0

278 - 279 = (-1)² = 1

Add them all

= 9 + 4 + 1 + 1 + 1 + 9 + 36 + 49 + 1 = 111

Varaince = 111/n - 1 (where n is number of samples)

= 111/12 - 1

= 111/11

= 10

Standard deviation = rootsquare of 10 = 3.16

First we will arrange the samples in ascending order

272, 276, 277, 278, 278, 278, 279, 279, 279, 280, 282, 285

Now mean is 279 and within 1 standard deviation will be between 275.84 - 282.16.

So, in this range 10 samples lie, so the percent of scores that will fall within one standard deviation = 10/12 × 100 = 83.33%

Hope u got it....