An insurance office buys paper by the ream, 500 sheets, for use in the copier, fax, and printer. Each ream lasts an average of 4 days, with standard deviation 1 day. The distribution is normal, independent of previous reams. a. Find the probability that the next ream out- lasts the present one by more than 2 days. b. How many reams must be purchased if they are to last at least 60 days with probability at least 80%

a) the probability that the next ream out- lasts the present one by more than 2 days is 0.0787

b) the number of reams that must be purchased is 19

Step-by-step explanation:

Given the data in the question;

Lets X₁ and X₂ be the two random variables that represents the first and second ream lasts

given that both random variables follow normal distribution with mean 4 and standard deviation 1.

a)

Find the probability that the next ream out- lasts the present one by more than 2 days.

P( X₂  - X₁ > 2 ) = P( [(X₂  - X₁ - E(X₂  - X₁)) / √(V(X₂  - X₁)] > [ (2-E(X₂  - X₁))/√(V(X₂  - X₁) ]

= 1 - P( Z ≤ [2-(μ₂  - μ₁)] / [√( V(X₂) + V(X₁) ) ] )

= 1 - P( Z ≤ [2-(4  - 4)] / [√( 1 + 1 )] )

= 1 - P = ( Z ≤ 2 / √2 )

= 1 - p( Z ≤ 1.4142 )

from excel; p( Z ≤ 1.41 ) = 0.9213

P( X₂  - X₁ > 2 ) = 1 - 0.9213

P( X₂  - X₁ > 2 ) = 0.0787

Therefore, the probability that the next ream out- lasts the present one by more than 2 days is 0.0787

b)

How many reams must be purchased if they are to last at least 60 days with probability at least 80%

total value is 4n

standard deviation is nσ

so

P(nX ≥ 60 ) = 0.80

P( nX-nμ /nσ  ≥  60-nμ/nσ) = 0.80

P( Z ≥  60-4n/n) = 0.80

1 - P( Z ≥  60-4n/n) = 0.80

P( Z < z ) = 0.20

now since z = 60-4n/n

from standard normal table

critical value of z corresponding to cumulative area of 0.20 is -0.841

so

z = - 0.841

60-4n/n =  -0.841

60 - 4n =  -0.841n

60 = -0.841n + 4n

60 = 3.159n

n = 60 / 3.159

n = 18.99 ≈ 19

Therefore, the number of reams that must be purchased is 19

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step-by-step explanation: