Question

The lifetime (in weeks) of printer toner cartridges is normally distributed with mean 4 and variance 1.5. A consumer buys a number of these cartridges with the intention of replacing them successively as soon as they become empty. The cartridges have independent lifetimes. The cartridges with lifetimes exceeding 5 weeks are classified as “satisfactory”.

(a) Find the approximate 95th percentile for the distribution of the life time of printer toner cartridges.

(b) What is the probability that a randomly selected cartridge is classified as “satisfactory”?

(c) If a sample of 20 cartridges is selected at random and the cartridges are inspected one by one, what is the probability that exactly 8 cartridges are classified as “satisfactory”?

(d) If a customer buys 30 cartridges, calculate the probability that at least 5 of the cartridges have not yet been used by the end of the 90th week.

Answer 1

a)

µ=   4                  
σ =    1.224744871                  
proportion=   0.95                  
                      
Z value at    0.95   =   1.64   (excel formula =NORMSINV(   0.95   ) )
z=(x-µ)/σ                      
so, X=zσ+µ=   1.64   *   1.224744871   +   4  
X   =   6.01   (answer)         

b)

µ =    4                  
σ =    1.224744871                  
                      
P ( X ≥   5.00   ) = P( (X-µ)/σ ≥ (5-4) / 1.22474487139159)              
= P(Z ≥   0.816   ) = P( Z <   -0.816   ) =    0.2071   (answer)

c)

Sample size , n =    20
Probability of an event of interest, p =   0.2071

	X	P(X)
P ( X = 8) = C (20,8) * 0.2071^8 * ( 1 - 0.2071)^12=	8	0.0263

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