The mean life of Speed- Hot geysers is normally distributed with a mean of 60 mo

ID: 3240177 • Letter: T

Question

The mean life of Speed- Hot geysers is normally distributed with a mean of 60 months and a standard deviation of 8 months.

1. For a randomly selected Speed - Hot geyser, what is the probability that the geyser will fail within 62 and 66 monhts?

2. What is the probability that a randomly selected Speed Hot geyser will fail within 4 years of the date of purchase?

3. Calculate the time period in which 60% of all geysers of this make will fail

4. Determine the guarantee period that must be set so that no more than 5% of geysers are replaced.

we are given that mean = 60 and sd = 8

1) For a randomly selected Speed - Hot geyser, what is the probability that the geyser will fail within 62 and 66 monhts?

so we first calculate the z scores as

z = (X-mean)/SD

(62-60)/8 <Z < (66-60)/8

0.25 <Z< 0.75

now to calculate the probability we shall use the z tables , please keep them handy

To find the probability of P (0.25<Z<0.75), we use the following formula:

P (0.25<Z<0.75 )=P ( Z<0.75 )P (Z<0.25 )

We see that P ( Z<0.75 )=0.7734

We see that P ( Z<0.25 )=0.5987.

At the end we have:

P (0.25<Z<0.75 )=0.1747

2 )

within 4 years means than 12*4 = 48 months

P(X<48), converting it into a z score as shown above

(48-60)/8 = -1.5 so we need to check

P(Z<-1.5), again from the z tables

P ( Z<1.5 )=1P ( Z<1.5 )=10.9332=0.0668

3. Calculate the time period in which 60% of all geysers of this make will fail

to calculate this we need to first find the z score that corresponds to 0.60

so we again check the z table and find the value as

0.2533

so using the z score formula

0.2533 = (X - 60)/8 , now solve for X

X = 0.2533*8 +60 = 62.064 months

no more than 5% geysers means that we need to check the value for p greater than 0.95

so the z value is 1.644

so using the z score formula

1.644 = (X - 60)/8 , now solve for X

X = 1.644*8 +60 = 73.15 months

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