SOLUTIONS ASSIGNMENT 1 STAT 231
NOTE: In the assignment, it was asked that a stem and leaf diagram of the data be prepared. This
diagram probably is not appropriate because of the limited range of the data, but for the record, is
shown below:
0| 4 5 6 8 8 8 9 9 9 9
1|000001111122222233455556
2.85
a. 2.855
b. 3.256
c. 2.940
d. The DM group had the largest standard deviation and therefore the most variation, while the
honey group had the smallest standard deviation and least variation. Note that all groups had
the same range.
2.101
a. We use Chebyshev’s Rule. At least 75% of the data for hand rubbers will lie within two standard
deviations of the mean, ie, in the interval (35-2(59) , 35+2(59). Since counts cannot be negative, this
gives the interval (0, 153).
b. In a similar manner we deduce that 75% of the data for hand washers will lie in the interval (0, 281).
c. Clearly, since the interval found for hand rubbers is smaller than that for hand washers, we infer that
hand rubbing is more effective.
2.123
a. z=2-> GPA = 2.7 +2 (0.5) = 3.7. In a similar manner we deduce:
z = -1 -> GPA = 1.7
z = 0.5 -> GPA = 2.95
z = -2.5 -> GPA = 1.45
b. GPA = 2.7 – 1.6(0.5) = 1.9
c. The Emprical Rule states that 68% of the GPA’s will lie within 1 standard deviation of the
mean. Thus, 68% will lie between 2.7 – 0.5 and 2.7 + 0.5, i.e., in the interval (2.2, 3.2). It
follows that 32% will lie outside this interval (either above or below) and therefore 16%
will lie above GPA 3.2. The z-score is 1.
The Empirical Rule also states that 95% will lie within 2 standard deviations of the mean.
Reasoning as we did above, it follows that 2.5% will lie above 2.7+2(0.5)=3.7. Therefore
Summa cum laud translates to a z-score of 2 and GPA 3.7. BIG NOTE: Must assume the
data is mound shaped and symmetric about the mean.