A student uses a pen for a certain time before he/she loses it or it runs out of

Question

A student uses a pen for a certain time before he/she loses it or it runs out of ink. It is found that the length of time the student uses a particular pen is governed by a x2 distribution with a mean of 9 days. The student likes to buy enough pens in August during the back-to-school sales so that he/she has enough pens to last the academic year which is 200 days. How many pens should the student buy in August so that the probability the student will not have to buy more pens during the academic year is at least 0.95? Would your answer change if the distribution of the time a student has the pen was a different distribution (say, an exponential distribution) with a mean of 9 days? Repeat the calculation for that case. Hint: If we assume that the number of pens will be large, argue that the length of time k pens will last should be well described by a Gaussian random variable. 6.

Explanation / Answer

Question 6

Here the pen operating time follows X2 distribution. so each pen lifetime follows that distribution with mean life = 9 days.

so,

as we know the property of chi- square distribution that  the sum of independent chi-squared variables is also chi-squared distributed. Specifically, if {Xi}i=1n are independent chi-squared variables with {ki}i=1n degrees of freedom, respectively, then Y = X1 + ? + Xn is chi-squared distributed with k1 + ? + kn degrees of freedom

so here k1 = k2 = k3 = 9

now we have to find the value of n where

CHIDIST (X2 > 200 ; 9n) > 0.05

so here by going trial and error, we get n = 19 here.

If the distribution is an exponential distribution. it can approximated to normal variable.

if there are minimum n such pens required.

Mean number of pens required = 9n

standard deviation of pens required = 9sqrt(n)

so,

Pr(X > 200 ; 9n ; 9 sqrt(n) ) > 0.05

Z = 1.645

(200 - 9n)/ [9sqrt(n)] = 1.645

200 - 9n = 14.805 * sqrt(n)

so n = 14 here.

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