BUSINESS STATISTICSCBEB1109FITRIAH MUNIRAH ALIASCEA140151Q6.17 (PG. 244)

TUTORIAL SLOT:WEDNESDAY 1PM-2PM

6.17 As player salaries have increased, the cost of attending baseball games has increased dramatically. The file BBCost 2010 contains the cost of four tickets, two beers, four soft drinks, four hot dogs, two game programs, two baseball caps, and the parking fee for one car for each of the 30 Major League Baseball teams in 2010:

115 121 127 132 141 151 158 160 161 162 168 170 172 173 178 180 184 207 208 212 216 217 221 222 227 227 250 316 330 335

Decide whether the data appear to be approximately normally distributed by a) comparing data characteristics to theoretical properties.

b) Constructing a normal probability plot

a. COMPARING DATA CHARACTERISTICS TO THEORETICAL PROPERTIES

Mean, µ = = = 194.70

Median = = = 15.5 (ranked value) = = 179

Mode = 227

Standard deviation

= 30 = 55.866

Interquartile Range = Q3 – Q1

Q3 = 3[ ] Q1 =

= 23.25 (ranked value) = 7.75≈ 23 ≈ 8= 221 = 160

Q3 – Q1= 221 – 160= 61

1.33 = 74.3018

Range = X largest – X smallest= 335 – 115= 220

6 = 6(55.866)= 335.196

The boxplot is skewed to the right. (The normal distribution is symmetric)

MEAN = 194.7MEDIAN = 179MODE = 227 The mean, median and mode have different

value. (In a normal distribution, the mean, median and mode are equal.)

IQR = 611.33 = 74.3018 The interquartile range is lower than 1.33. (In

a normal distribution, the interquartile range is 1.33 standard deviations.)

RANGE = 2206 = 335.196 The range is much lower than 6. (In

a normal distribution, the range is 6 standard deviations.)

b. CONSTRUCTING A NORMAL PROBABILITY PLOT

Find the cumulative probability by using the formula above

Use NORM.S.INV to get the Z-score (inverse)

Select Z-score for X valuesSorted data for Y values

CONCLUSIONThe normal probability plot shows

that the data does not appear to be approximately normally distributed.

Overall, this data set greatly differs from the theoretical properties of the normal distribution.