Question

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%

Answer 1

The probability that the next ream outlasts the present one by more than 2 days is approximately 0.0668. To last at least 60 days with a probability of at least 80%, 5 reams must be purchased.

Step-by-step explanation:

In this problem, we have a normal distribution with a mean of 4 days and a standard deviation of 1 day for the duration of each ream of paper.

a. To find the probability that the next ream outlasts the present one by more than 2 days, we need to find the area under the normal curve to the right of 2. We can use a standard normal distribution table or a calculator to find this probability. Using a calculator, the probability is approximately 0.0668.

b. To determine how many reams must be purchased to last at least 60 days with a probability of at least 80%, we need to find the number of standard deviations required to reach this duration. We can use the formula z = (x - μ) / σ, where x is the desired duration, μ is the mean, and σ is the standard deviation. Rearranging the formula to solve for x, we have x = z * σ + μ. Using a standard normal distribution table or a calculator, we can find the z-score corresponding to a probability of at least 80% (0.8), which is approximately 0.8416. Substituting the values into the formula, we have x = 0.8416 * 1 + 4. The number of reams required to last at least 60 days is approximately 4.8416, which rounds up to 5 reams.

Answer 2

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