n =15
∑X = 99-15.5 = 83.5
so
=∑X/n = 83.5/15 = 5.57=∑X/n = 99/16 = 6.19
Without the 15.5:
n =15
∑X = 99-15.5 = 83.5
so =∑X/n = 83.5/15 = 5.57
Problem 2:
The ordered dataset is:
3.0 3.5 3.5 4.0 4.5 4.5 5.5 5.5 6.0 6.0 6.5 6.5 7.5 8.0 9.0 15.5
Now
Q1=P25, np/100 = 16·25/100 = 4, 4th observation is 4
P67, np/100 = 16·70/100 = 11.2, round up to 12, 12th observation is 6.5
Problem 3:
IQR = Q3-Q1 = 24-16 = 8
LF = Q1-1.5·IQR = 16-1.5·8 = 4 < Min
so left whisker goes to Min
UF = Q3+1.5·IQR = 24+1.5·8 = 36 < Max
so right whisker goes to 36 and all larger observations are stars:
Problem 4:
a) =1250 s=0.75
Perfectly possible
(example: data are the lengths of wooden beams made by a machine, all are almost exactly 1250 inches)
b) =12.0 median=23.5
Perfectly possible
(example: data are number of cigarettes smoked in a week, then we have many 0's (non-smokers) but also many high numbers. If more than half the people in the sample smoke we might get these numbers)
c) s=250 range=300
Possible but unlikely, range/4 (=75) should be close to s.
(can't think of an example, after all it's unlikely, but not mathematically impossible)
d)
| Minimum | Q1 | Median | Q3 | Maximum |
| -15 | -4 | 7 | 16 | 16 |
e)
| Minimum | Q1 | Median | Q3 | Maximum |
| 10 | 22 | 57 | 45 | 99 |
f) =-5.1 s=-0.5
Not possible s≥0 always [s=√(Sxx/(n-1)) ]
Problem 5
a) Find the mean of the incomes
Stat > Basic Statistics > Display Discriptive Statistics > Variable: Income
b) Find the mean of the incomes of the females
Stat > Basic Statistics > Display Discriptive Statistics > Variable: Income, By Variable: Gender
c) Find the ID number of the employee with the highest income
Data > Sort > Sort data in columns : ID, by variable: Income
d) Find the percentage of employees with a Satisfaction of 5
Stat > Tables > Tally Individuel variables > Variable: Satisfaction, check percentage
e) Use the Calculator command to find the number of employees with a satisfaction of 4 or 5
Calc > Calculator, Store in: c8 Expression: SUM('Satisfaction' >= 4 )
