Tutorial on Section 4 SPECIAL DISTRIBUTIONS (continuous distributions)
1. LetX∼N(μ,σ2).
(a) Show, using the mgf of Z = (X − μ)/σ, that the third, fourth and fifth central
momentsofX aregivenbyE[(X−μ)3]=0,E[(X−μ)4]=3σ4,E[(X−μ)5]=0.
(b) Find an expression for E[(X − μ)6].
2. The lifetime in hours, X, of a particular type of electron tube is distributed as an ex- ponential random variable with mean 100 hours, so f(x) = 0.01e−0.01x, x > 0. Five of these tubes are used in a particular device and the device functions only so long as all five tubes are working. Assume the necessary independence, and let Y be the lifetime of such a device.
(a) Show that P (Y > y) = e−0.05y and hence identify the distribution of Y .
(b) Find the probability that, of three such devices operating independently, at least
two will fail within 40 hours of operation. [0.950]
(c) Find the median lifetime of (i) a tube, and (ii) a device. [i) 69.3 hrs, ii) 13.9 hrs]
3. Every week a salesman has to go from his office to collect his week’s supplies from one of two depots and then return to his office. From experience he knows that the time it takes to go from his office to depot A or back is normally distributed with mean 25 and variance 3, and the time it takes to obtain his supplies at depot A is independently normally distributed with mean 15 and variance 10. For depot B, the travelling time is normally distributed with mean 22 and variance 11, and the suppy time is independently normally distributed with mean 20 and variance 14. The time units are minutes.
(a) What is the probability that the salesman can go to depot A, collect his supplies, and return to his office (‘do the round trip’) in less than 75 minutes? [0.9938]
(b) Which depot should he use if he takes the route which allows him to do the round trip in the smaller expected time?
(c) Which depot should he use if he wants to maximise the chance that he can do the round trip in under 75 minutes?
(d) What is the probability that on any day the round trip would take less time using depot A than depot B? [0.4443]
Notes: (i) a linear combination of independent normal random variables is itself a normal random variable; (ii) E[X + Y ] = E[X] + E[Y ]; (iii) if X and Y are independent then Var[X + Y ] = Var[X] + Var[Y ].
A customer buys 5 mm ball bearings from a manufacturer. The specification calls for bearings with diameters within the range 5.00±0.06 mm. The manufacturing process yields bearings whose diameters can be assumed to follow a normal distribution with mean μ and standard deviation σ.
In the manufacturer’s inspection process a total of 12% of the output is rejected, 1.5% being smaller than the specified lower limit and 10.5% being larger than the specified upper limit.
Find the values of μ and σ (to 3 dp). [μ = 5.016mm, σ = 0.035mm] 1

(b) It costs 5p to produce and inspect each ball bearing. Satisfactory bearings are sold to the customer for 8p each, while bearings which fail to meet the specifications have scrap value of 0.5p each.
Find the manufacturer’s expected profit on a production batch of 100 000 ball bear- ings. [£2100]
(c) When the customer receives the satisfactory bearings he submits them to a further inspection process which sorts them into three categories: grade A for precision work with diameters in the range 5.00 ± 0.01 mm; grade B (small) with diameters less than 4.99 mm; and grade C (large) with diameters greater than 5.01 mm.
What percentage of the bearings the customer receives are classified into each of the three grades? [A 23.1%, B 24.4 %, C 52.7 %]
(d) The manufacturer would like to increase his profit on such a batch. It is possible
(i) to eliminate the bias in the machine setting (so that the mean diameter of man- ufactured bearings is 5.00 mm) without altering the variability of the diameters, at a cost of £100;
(ii) to eliminate the bias in the machine setting and reduce the standard deviation of the diameters to 0.025 mm, at a cost of £800.
Determine which of these procedures, if either, will increase the manufacturer’s ex- pected profit, and hence state the maximum profit he can expect to make on a batch. [£2246]

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