9/ 07/2014 The probability distribution for the number of repairs, N, a brand of refrigerator requires over a 5-year period is:
N |
0 |
1 |
2 |
3 |
4 |
5 |
6 |
P(N) |
0.17 |
0.32 |
0.23 |
0.15 |
0.08 |
0.04 |
0.01 |
a)
Estimate the percentile corresponding to 3 repairs. Interpret
it,
in context.
b)
Determine the 3 quartiles. Then, calculate and interpret
the
IQR, in context.
c) Which measures
of
Centre ought to be used to accurately describe the number of repairs
needed over a 5-year period? Explain.
d)
Do you expect the Mean number of repairs to be higher or lower than
the Median? Why?
e) Calculate the expected number of repairs
needed over a 5-year period.
f) Calculate the s.d. of the number
of repairs needed over a 5-year period. Interpret
it,
in context.
Scroll down for the solution.
9/07/2014
Solution.
a)
P(X <
3)
= P(X = 0) + ...P(X = 3) = 0.17 + … 0.15 = 0.87.
The 87^{th}
percentile
being 3 repairs indicates that 87% of repairs over a 5-year period
were 3 or less or
better still
The
87^{th}
percentile
being 3 repairs indicates that 87% of refrigerators needed / had 3 or
less over a 5-year period. Observe
the detailed context!
b)
Q1 = 1 since P(X <
1)
>
0.25
and P(X >
1)
>
0.75.
<------------- This
is the technical definition of percentile, p,
corresponding
to X = a:
P(X
<
a)
>
p
and
P(X >
a)
>
1
– p...and
this definition permits us to “cross” the percentage!
Q2
= 2 since P(X <
2)
>
0.50
and P(X >
2)
>
0.50.
Q3
= 3 since P(X <
3)
>
0.75
and P(X >
3)
>
0.25.
IQR
= 3 – 1 = 2 is the spread
or
range
of
the middle 50% of the number of refrigerator repairs needed in a
5-year period. Observe
the detailed context!
c)
Median and Mode since since this a right-skewed distribution.
d)
The Mean shall likely be higher than the Median since it is a
right-skewed distribution, so the Mean shall be “pulled up”
by the extreme values of X on the upper-end. Observe
the detailed context!
e)
E(X) ≈
Average
= ΣX·P(X) = 0·0.17 + ...6·0.15 = 1.81
using 1-Var Stats L1, L2 command on the graphing calculator with
X-values in L1, and probabilities in L2.
f) σ(X) = 1.3978
using 1-Var Stats L1, L2 with X-values in L1, and probabilities in
L2.
BE
FAMILIAR WITH ALL
INTERPRETATIONS!
The s.d. of 1.3978 is a measure of the average variability of the different / each number of repair(s) from the mean number of repairs of 1.81 Observe the detailed context! OR
The s.d. of 1.3978 indicates that, on average, the different / each number of repair(s) is about 1.3978 from the mean number of repairs of 1.81Observe the detailed context! OR
The s.d. of 1.3978 indicates that, on average, the different / each number of repair(s) differs from the mean number of repairs of 1.81 by about 1.3978 Observe the detailed context! OR
The s.d. of 1.3978 indicates that, on average, the difference between the different / each number of repair(s) and the mean numbers of repair of 1.81 is about 1.3978. Observe the detailed context!