In my last post I discussed how to use the normal distribution function already added in MS Excel. In this post I will continue with the examples and we will how we use them. In this post i will not discuss the manual solution of the problem but only the Excel Based solution.
The following problems are taken from Introduction to Statistics By Walpole, 13th Edition Chapter 07, Page 197-200.
I wil elaborate how you can use excel formula to calculate the probabilites.
4. A soft-drink machine is regulated so that it discharges an average of 200 milliliters per cup. If the amount of drink is normally distributed with a standard deviation equal to 15 milliliters:
(a) What fraction of the cups will contain more than 224 milliliters?
(b) What is the probability that a cup contains between 191 and 209 milliliters?
(c) How many cups will likely overflow if 230-mililiters cups are used for the next 1000 drinks.
(d) Below what value do we get the smallest 25% of the drinks?
Solution:
Here $\mu$ = 200 mililiters and $\sigma$ = 15 mililiters
(a) =$1-Normdist(224,200,15,1) = 0.0548
(b) =Normdist(209,200,15,1)-Normdist(191,200,15,1) = 0.451
(c) =(1-Normdist(230,200,15,1))*1000 = 22.75
(d) =Norminv(0.25,200,15) = 189.89
9. The heights of 1000 students are normally distributed with a mean of 174.5 cm and a std. deviation of 6.9 cm. Assuming that he heights recorded are recorded to the nearest half of a centimeter, how many of these students would you expect to have heights,
(a) Less than 160.0 cm?
(b) Between 171.5 and 182.0 cm inclusive?
(c) Equal to 175.0 cm?
(d) Greater than or equal to 188.0 cm?
Solution:
Here $\mu$ = 174.5 cm mililiters and $\sigma$ = 6.9 cm
(a) =1000*Normdist(160,174.5,6.9,1)
(b) =(Normdist(182.5-0.5,174.5,6.9,1)-Normdist(171.5-0.5,174.5,6.9,1))*1000
(c) =(Normdist(175+0.5,174.5,6.9,1)-Normdist(175-0.5,174.5,6.9,1))*1000
(d) =(1-Normdist(188-0.5,174.5,6.9,1))*1000
Note: In (b), (c) and (d) we substracted and added 0.5 from values of $x$ to make them inclusive. We are converting point estimate into an intervel estimate through this operation.
16. The average life of a certain type of small motor in 10 years, with a stad. Deviation of 02 years. The manufacturer replaces free all motors that fail while under guarantee. If he is willing to replace only 3% of the motors that fail, how long a guarantee should he offer? Assume that the lives of the motors follow a normal distribution.
Solution:
Here $\mu$ = 10 years and $\sigma$ = 2 years and $\phi$ = 0.03
=Norminv(0.03,10,2) = 6.238 years
The statistical concepts are self-explanatory. I hope that you are finding these posts helpful certain way. Hope to listen your feedback soon. Take Care
The following problems are taken from Introduction to Statistics By Walpole, 13th Edition Chapter 07, Page 197-200.
I wil elaborate how you can use excel formula to calculate the probabilites.
4. A soft-drink machine is regulated so that it discharges an average of 200 milliliters per cup. If the amount of drink is normally distributed with a standard deviation equal to 15 milliliters:
(a) What fraction of the cups will contain more than 224 milliliters?
(b) What is the probability that a cup contains between 191 and 209 milliliters?
(c) How many cups will likely overflow if 230-mililiters cups are used for the next 1000 drinks.
(d) Below what value do we get the smallest 25% of the drinks?
Solution:
Here $\mu$ = 200 mililiters and $\sigma$ = 15 mililiters
(a) =$1-Normdist(224,200,15,1) = 0.0548
(b) =Normdist(209,200,15,1)-Normdist(191,200,15,1) = 0.451
(c) =(1-Normdist(230,200,15,1))*1000 = 22.75
(d) =Norminv(0.25,200,15) = 189.89
9. The heights of 1000 students are normally distributed with a mean of 174.5 cm and a std. deviation of 6.9 cm. Assuming that he heights recorded are recorded to the nearest half of a centimeter, how many of these students would you expect to have heights,
(a) Less than 160.0 cm?
(b) Between 171.5 and 182.0 cm inclusive?
(c) Equal to 175.0 cm?
(d) Greater than or equal to 188.0 cm?
Solution:
Here $\mu$ = 174.5 cm mililiters and $\sigma$ = 6.9 cm
(a) =1000*Normdist(160,174.5,6.9,1)
(b) =(Normdist(182.5-0.5,174.5,6.9,1)-Normdist(171.5-0.5,174.5,6.9,1))*1000
(c) =(Normdist(175+0.5,174.5,6.9,1)-Normdist(175-0.5,174.5,6.9,1))*1000
(d) =(1-Normdist(188-0.5,174.5,6.9,1))*1000
Note: In (b), (c) and (d) we substracted and added 0.5 from values of $x$ to make them inclusive. We are converting point estimate into an intervel estimate through this operation.
16. The average life of a certain type of small motor in 10 years, with a stad. Deviation of 02 years. The manufacturer replaces free all motors that fail while under guarantee. If he is willing to replace only 3% of the motors that fail, how long a guarantee should he offer? Assume that the lives of the motors follow a normal distribution.
Solution:
Here $\mu$ = 10 years and $\sigma$ = 2 years and $\phi$ = 0.03
=Norminv(0.03,10,2) = 6.238 years
The statistical concepts are self-explanatory. I hope that you are finding these posts helpful certain way. Hope to listen your feedback soon. Take Care
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