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 measure
s 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 87th 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!