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In: Statistics and Probability

I need to answer these questions in rstudio - Continuous Probability Distributions Question 4: A printer...

I need to answer these questions in rstudio - Continuous Probability Distributions

Question 4: A printer manufacturer determines that the time to failure of one of their ink cartridges is an approximately normally distributed random variable with a mean of 5570 pages and a standard deviation of 237 pages. For marketing purposes they wish to offer a warranty to purchasers, but to protect profitability of the company, they wish to set the warranty period so that at most 2% of the cartridges fail under warranty. What is the longest warranty period they can offer?

Question 5: When X1 and X2 are both normally distributed random variables with means µ1 and µ2, and variances σ1 2 and σ2 2 , respectively, their sum, X1 + X2 is also a normally-distributed random variable and has the mean value µ1 + µ2 and the variance σ1 2 + σ2 2 . Suppose Hank is going on a long trek, and he leaves with two different but fully charged cell phone batteries. Both batteries have lifetimes (on standby) which are from populations which are approximately normally-distributed. The first battery comes from a population with a mean of 65.24 hours and a standard deviation of 7.42 hours, and the second battery comes from a population with a mean lifetime of 51.22 hours and a standard deviation of 10.67 hours. Hank`s plan is to use one battery until it dies, and then use the second one, in order to prolong his phone usage. Compute the probability that a) Hank will get at least 120 hours of phone usage out of his two batteries b) Hank will get less than 100 hours of phone usage out of his two batteries

Solutions

Expert Solution

4)

            Let X denote failure time of cartridge, then it is given than X follows normal distribution with mean 5570 pages and standard deviation 237

We need to calculate the point ‘C’ at which Probability of failure of the cartridge below C is 2%

i.e P(X < C ) = 0.02

For this we are going to use qnorm() command in R studio.

the syntax is

qnorm(p, mean = 0, sd = 1, lower.tail = TRUE)

Now executing the command we get

i.e. only 2% cartridges are expected fall below 5083.262 pages. Hence the answer.

5)

            Let X = X1+X2 denote the total hours of phone usage

            Given that, X1 follows normal with mean 65.24 and SD 7.42 hrs and

            X2 follows Normal with mean 51.22 and SD 10.67 hrs.

            So X= X1+X2 follows normal with mean 65.24 + 51.22 = 116.46

            Variance of X= Variance of X1 + Variance of X2 =7.422 + 10.672 = 168.91

            SD (X) = square root of (168.91) =12.99

  1. Now, the probability that Hank will get at least 120 hours of phone usage out of his two batteries is = P( X >120)

for this we will use pnorm() function. the syntax is

pnorm(q, mean = 0, sd = 1, lower.tail = FALSE)

we set lower.tail =FALSE because here we need the upper tail probability. After executing the command we get the result.

so the required probability is 0.3988399

  1. The probability of Hank will get less than 100 hours of phone usage out of his two batteries is P( X <120)
    for this we will use pnorm() function. the syntax is

pnorm(q, mean = 0, sd = 1, lower.tail = TRUE)

Here we set lower.tail = TRUE since we need the lower probability

The output will get is


The required probaility is 0.0996944



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