Answer:
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
