18 Normal Distribution Problem Set
NORMAL DISTRIBUTION | PRACTICE PROBLEMS
Complete the following to reinforce your understanding of the concept covered in this module.
PROBLEM 1: The area under a standard normal curve with z-scores of π§ = 0 πππ π§ = 0.78 is most close to: A. 0.19 B. 0.29 C. 0.39 D. 0.49
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SOLUTION 1: The NORMAL DISTRIBUTION TABLE can be referenced under the subject of ENGINEERING PROBABILITY AND STATISTICS on page 9.4 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. In this problem, we are asked to find the area under a standard normal curve given a set up z-scores. When we hear the statement βarea under a curveβ, we might be quick to assume this will be an integrationβ¦but itβs not, and in fact, no correct answer can be derived taking this route since there is no function to differentiate against. Knowing the triggered terminology used in the problem statement shows us exactly what they are looking for. First, a standard normal curve is simply a bell curve, a simple probability distribution that is tabulated in the Unit Normal Distribution tables given to us in our NCEES Reference Handbook. So thatβs the first thing, when you hear standard normal curve, we are working a Normal Distribution problem. Second, a z-score is short for what is known as a Standard Score. Donβt know where the βzβ comes from, but thatβs how it has evolved. A Standard Score is directly related to Statistics, and more specifically, Normal Distributions. When we see this term used in
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a problem statement, let it trigger in our mind once again that we are working a Normal Distribution problem. Given any z-score we are able to reference our Unit Normal Distribution tables and determine a probability of that standard score occurring based on the scenario in which we are working. In this problem, we are asked to determine what is referred to as a Banded Probabilityβ¦between a z-score of 0 and up to a z-score of .78. Considering these two probabilities, we will define them both and then combine the to develop a cumulative probability, which will be our total probability represented by the expression: Total Probability = π π₯ > 0 + π π₯ < .78 So this is saying that we will add the probability that a z-score will be greater than 0 plus a z-score of x being less than .78. But here is where some tweaking it going to come in to play. Our Unit Normal Distribution table will not present to us this exact case where the probability is banded between two non uniform z-scores. We will have to get creative in taking the values that they do define and play around with them a bit to get our answer. Letβs illustrate this.
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When we hop back to our Normal Distribution Tables, the first thing we will notice is that there is no column that refers to a z-scoreβ¦at least explicitly. This can be confusing, but the z-score is indeed representedβ¦in the x column, so this is where we will reference each z-score. Running to the right, we see that there are a number of different scenarios which we are presented, in each case, we will choose one and run with itβ¦adjusting it as needed knowing that the Cumulative Probability always adds up to 1 under a Normal Distribution Curve. In this problem, we will be using the R(x) column Flipping back to our tables and starting with z = 0, we can highlight the data to the right up to the π (π₯) column. The single value we reference when the row and the column line up will be the Probability of this scenario occurring:
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This tells us that for a z-score of 0, we have: π 0 = 0.5 But more importantly, we have to expand on that in relation to the column in which we are referencing. The R(x) column gives us a probability from our z-score, x, to infinite, therefore, we can conclude that: π π₯ > 0 = 0.5 This is be used as the basis for the lower band of the probability we are looking for. Moving to a z-score of .78, we can highlight the data to the right up to the π (π₯) column. The single value we reference when the row and the column line up will be the Probability of this scenario occurring:
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This tells us that for a z-score of .78, or .8 because itβs the closest, we have: π . 8 = 0.2119 Again, we have to expand on that context of this probability so that we can correctly apply it to the cumulative probability we are looking for. The R(x) column gives us a probability from our z-score, x, to infinite, therefore, we can conclude that: π π₯ > .8 = 0.2119 We can now bring these two values to determine what the cumulative probability is between the lower and upper band that we have been given. Since the value we found for the z-score of 0 included all value up to infinite, which is past our second z-score of .8β¦we need to account for this. This wonβt be a problem because the value we got for our second z-score of .78 is for all values above that z-score. Therefore, to get the total banded probability, we will take the complete probability for the z-score of 0 up to infinite (our lower band) and subtract the complete probability for our second z-score of .78 up to infinite (our upper band), leaving us the probability between our lower band and the upper band. Putting this in to an expression, this can then be written as:
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Total Probability = π π₯ > 0 β π π₯ > .78 Or rather: π 0 < π₯ < .78 = π π₯ > 0 β π π₯ > .78 Having defined: π π₯ > 0 = .5 π π₯ > .78 = .2119 We plug in our values and get: π 0 < π₯ < .78 = .5 β .2119 Or that the area under the curve, the probability, is: π 0 < π₯ < .78 = .5 β .2881
Therefore, the correct answer is B. 0.29
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PROBLEM 2: The area under the standard normal curve with z-scores of π§ = β0.43 πππ π§ = 0.78 is most close to: A. 0.2984 B. 0.3724 C. 0.4435 D. 0.4721
SOLUTION 2: The NORMAL DISTRIBUTION TABLE can be referenced under the topic of ENGINEERING PROBABILITY AND STATISTICS on page 9.4 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. We are given two z-scores that we can use to calculate the area under the standard normal curve. It is important to remember that we not able to use integration here, so we need to use the normal distribution tables to calculate the probability for each zscore. We then look at the normal distribution tables to find the corresponding probability values for each z-score.
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We use the F(x) column as we are looking to calculate the area under the curve from ββ π‘π π₯, π€βπππ π₯ = β0.43.
We use the πΉ(π₯) column to indicate πΉ π₯ < π₯ and π₯ = 0 row. We find the area under the curve as π π₯ > 0 = π 0.4 = 0.6554 .
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We use the π (π₯) column, as we are looking to calculate the area under the curve from π₯ π‘π β, π€βπππ π₯ = 0.78.
We use the π (π₯) column to indicate π 0.78 > 0 and π₯ = 0.8 row. We find the area under the curve as π 0.78 > 0 = π 0.8 = 0.2119. We then sum the probabilities of each z-score to calculate the area under the curve: π΄πππ πππππ πΆπ’ππ£π = 0.6554 β 0.2119 = 0.4435
Therefore, the correct answer is π. π. ππππ
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PROBLEM 3: It was found that the mean length of 100 parts produced by a machine was 20.05 mm with a standard deviation of 0.02 mm. The probability that a part selected at random would have a length between 20.06 mm and 20.07 mm is most close to: A. 12.11% B. 14.98% C. 17.30% D. 21.48%
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SOLUTION 3: The formula for CALCULATING A Z-SCORE can be referenced under the topic of NORMAL DISTRIBUTION (GAUSSIAN DISTRIBUTION) on page 39 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. Every normal random variable π₯ can be transformed into a z-score using the equation below:
π=
π₯βπ π
Where: β’ π₯ is the normal random variable β’ π is the mean β’ π is the standard deviation of π₯ To analyze this problem, we will calculate the z-score and then use the normal distributions tables, to find probabilities associated with the z-scores. We are given the following values to solve for z-scores of each wage range: β’ x] = 20.06 β’ π₯^ = 20.07 β’ π = 20.05 β’ π = 0.02
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For the first group of with a mean length of 20.06 mm, we will calculate the z-score, π] , of the probability a wage falls in that range as:
π§] =
(20.06 β 20.05) = 0.5 0.02
For the second group of with a mean length of 20.07 mm, we will calculate the z-score, π^ , of the probability a wage falls in that range as:
π§^ =
(20.07 β 20.05) =1 0.02
Which means that: π 20.06 < π₯ < 20.07 = π(0.5 < π§ < 1) We then look at the normal distribution tables to find the corresponding probability values for each z-score. A unit normal distribution table is included at the end of the reference handbook section on probability and statistics. In the table, the following notations are utilized: β’ πΉ π₯ = π‘βπ ππππ π’ππππ π‘βπ ππ’ππ£π ππππ β β π‘π π₯ β’ π π₯ = π‘βπ ππππ π’ππππ π‘βπ ππ’ππ£π ππππ π₯ π‘π β β’ π π₯ = π‘βπ ππππ π’ππππ π‘βπ ππ’ππ£π ππππ β π₯ π‘π π₯ β’ πΉ βπ₯ = 1 β πΉ(π₯)
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The NORMAL DISTRIBUTION TABLE can be referenced under the topic of ENGINEERING PROBABILITY AND STATISTICS on page 9.4 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. For the first group of with a mean length of 20.06 ππ, we use the z-score, π] = 0.5, to find the probability the wages are GREATER THAN 20.06 ππ. We use the π(π₯) column to indicate π π₯ > 20.06 and π₯ = 0.5 row, as this is the closet value we can use without interpolating. We use the πΉ(π₯) column as we are looking to calculate the area under the curve from π₯ π‘π β, π€βπππ π₯ = 0.5.
π π₯ > 20.06 = π(0.5) = 0.3829
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For the second group of with a mean length of 20.07 mm, we use the z-score, π^ = 1, to find the probability the wages are LESS THAN 20.07 mm. We use the π(π₯) column to indicate π π₯ < 20.07 and π₯ = 1 row. We use the W(x) column as we are looking to calculate the area under the curve from ββ π‘π π₯, π€βπππ π₯ = 20.07.
π π₯ > 20.07 = π 1 = 0.6827
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We then sum the probabilities of each range to calculate the probability that a part selected at random would have a length between 20.06 mm and 20.07 mm. As we are accounting for the under the curve a factor of 2, we need to divide each of the probabilities by a factor of 2: π π₯ < 20.07 π π₯ > 20.06 0.6827 0.3829 β = β = 0.1498 = 14.98% 2 2 2 2
Therefore, the correct answer is π. ππ. ππ%
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PROBLEM 4: The average life of a certain type of engine is 10 years, with a standard deviation of 2 years. If the manufacturer is willing to replace only 3% of the motors that fail, how long should the guarantee be? Assume the motors follow a normal distribution. A. 5.70 π¦ππππ B. 6.20 π¦ππππ C. 6.75 π¦ππππ D. 7.25 π¦ππππ
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SOLUTION 4: The FORMULA FOR CALCULATING A Z-SCORE can be referenced under the topic of NORMAL DISTRIBUTION (GAUSSIAN DISTRIBUTION) on page 39 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. In this problem we need to work backwards to determine the value (in years) that will give the bottom 3% of the distribution. These are the motors that the manufacturer is willing to replace under the guarantee. To analyze this problem, knowing it is normally distribution, we will use the normal distribution able to find the area under the normal distribution curve. We are concerned with the area under the normal distribution curve that is reflected by the 3% that the manufacturer willing to replace.
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As we know the area under the curve, we can work backwards to use the normal distribution table to determine the z-score associated with the probability of 97%. We use the πΉ(π₯) column as we are looking to calculate the area under the curve from π₯ π‘π β, π€βπππ π₯ = 0.9713, as this is the closest value to 0.97 or 97% in the πΉ(π₯) column.
We find that for π (0.97) the corresponding z-score is π§ = β1.9. Every normal random variable π₯ can be transformed into a z-score using the equation below:
π=
π₯βπ π
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Where: β’ π₯ is the normal random variable β’ π is the mean β’ π is the standard deviation of π₯ To analyze this problem, we will calculate the z-score and then use the normal distributions tables, to find probabilities associated with the z-score. We are given the following values to solve for the z-score: β’ x] = πΏπππ ππ πππ‘ππ β’ π₯^ = πΊπ’πππππ‘ππ ππππππ β’ π = 10 β’ π=2 Plugging in the known values and calculated z-score, we can solve for the number of years for the guarantee period:
β1.9 =
π₯ β 10 2
Solving for π₯, we find the guarantee period should be 6.20 π¦ππππ .
The correct answer is π. π. ππ π²πππ«π¬
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Complete the following to reinforce your understanding of the concept covered in this module.
PROBLEM 1: The area under a standard normal curve with z-scores of π§ = 0 πππ π§ = 0.78 is most close to: A. 0.19 B. 0.29 C. 0.39 D. 0.49
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SOLUTION 1: The NORMAL DISTRIBUTION TABLE can be referenced under the subject of ENGINEERING PROBABILITY AND STATISTICS on page 9.4 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. In this problem, we are asked to find the area under a standard normal curve given a set up z-scores. When we hear the statement βarea under a curveβ, we might be quick to assume this will be an integrationβ¦but itβs not, and in fact, no correct answer can be derived taking this route since there is no function to differentiate against. Knowing the triggered terminology used in the problem statement shows us exactly what they are looking for. First, a standard normal curve is simply a bell curve, a simple probability distribution that is tabulated in the Unit Normal Distribution tables given to us in our NCEES Reference Handbook. So thatβs the first thing, when you hear standard normal curve, we are working a Normal Distribution problem. Second, a z-score is short for what is known as a Standard Score. Donβt know where the βzβ comes from, but thatβs how it has evolved. A Standard Score is directly related to Statistics, and more specifically, Normal Distributions. When we see this term used in
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by Prepineer | Prepineer.com
a problem statement, let it trigger in our mind once again that we are working a Normal Distribution problem. Given any z-score we are able to reference our Unit Normal Distribution tables and determine a probability of that standard score occurring based on the scenario in which we are working. In this problem, we are asked to determine what is referred to as a Banded Probabilityβ¦between a z-score of 0 and up to a z-score of .78. Considering these two probabilities, we will define them both and then combine the to develop a cumulative probability, which will be our total probability represented by the expression: Total Probability = π π₯ > 0 + π π₯ < .78 So this is saying that we will add the probability that a z-score will be greater than 0 plus a z-score of x being less than .78. But here is where some tweaking it going to come in to play. Our Unit Normal Distribution table will not present to us this exact case where the probability is banded between two non uniform z-scores. We will have to get creative in taking the values that they do define and play around with them a bit to get our answer. Letβs illustrate this.
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When we hop back to our Normal Distribution Tables, the first thing we will notice is that there is no column that refers to a z-scoreβ¦at least explicitly. This can be confusing, but the z-score is indeed representedβ¦in the x column, so this is where we will reference each z-score. Running to the right, we see that there are a number of different scenarios which we are presented, in each case, we will choose one and run with itβ¦adjusting it as needed knowing that the Cumulative Probability always adds up to 1 under a Normal Distribution Curve. In this problem, we will be using the R(x) column Flipping back to our tables and starting with z = 0, we can highlight the data to the right up to the π (π₯) column. The single value we reference when the row and the column line up will be the Probability of this scenario occurring:
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This tells us that for a z-score of 0, we have: π 0 = 0.5 But more importantly, we have to expand on that in relation to the column in which we are referencing. The R(x) column gives us a probability from our z-score, x, to infinite, therefore, we can conclude that: π π₯ > 0 = 0.5 This is be used as the basis for the lower band of the probability we are looking for. Moving to a z-score of .78, we can highlight the data to the right up to the π (π₯) column. The single value we reference when the row and the column line up will be the Probability of this scenario occurring:
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This tells us that for a z-score of .78, or .8 because itβs the closest, we have: π . 8 = 0.2119 Again, we have to expand on that context of this probability so that we can correctly apply it to the cumulative probability we are looking for. The R(x) column gives us a probability from our z-score, x, to infinite, therefore, we can conclude that: π π₯ > .8 = 0.2119 We can now bring these two values to determine what the cumulative probability is between the lower and upper band that we have been given. Since the value we found for the z-score of 0 included all value up to infinite, which is past our second z-score of .8β¦we need to account for this. This wonβt be a problem because the value we got for our second z-score of .78 is for all values above that z-score. Therefore, to get the total banded probability, we will take the complete probability for the z-score of 0 up to infinite (our lower band) and subtract the complete probability for our second z-score of .78 up to infinite (our upper band), leaving us the probability between our lower band and the upper band. Putting this in to an expression, this can then be written as:
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Total Probability = π π₯ > 0 β π π₯ > .78 Or rather: π 0 < π₯ < .78 = π π₯ > 0 β π π₯ > .78 Having defined: π π₯ > 0 = .5 π π₯ > .78 = .2119 We plug in our values and get: π 0 < π₯ < .78 = .5 β .2119 Or that the area under the curve, the probability, is: π 0 < π₯ < .78 = .5 β .2881
Therefore, the correct answer is B. 0.29
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PROBLEM 2: The area under the standard normal curve with z-scores of π§ = β0.43 πππ π§ = 0.78 is most close to: A. 0.2984 B. 0.3724 C. 0.4435 D. 0.4721
SOLUTION 2: The NORMAL DISTRIBUTION TABLE can be referenced under the topic of ENGINEERING PROBABILITY AND STATISTICS on page 9.4 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. We are given two z-scores that we can use to calculate the area under the standard normal curve. It is important to remember that we not able to use integration here, so we need to use the normal distribution tables to calculate the probability for each zscore. We then look at the normal distribution tables to find the corresponding probability values for each z-score.
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We use the F(x) column as we are looking to calculate the area under the curve from ββ π‘π π₯, π€βπππ π₯ = β0.43.
We use the πΉ(π₯) column to indicate πΉ π₯ < π₯ and π₯ = 0 row. We find the area under the curve as π π₯ > 0 = π 0.4 = 0.6554 .
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We use the π (π₯) column, as we are looking to calculate the area under the curve from π₯ π‘π β, π€βπππ π₯ = 0.78.
We use the π (π₯) column to indicate π 0.78 > 0 and π₯ = 0.8 row. We find the area under the curve as π 0.78 > 0 = π 0.8 = 0.2119. We then sum the probabilities of each z-score to calculate the area under the curve: π΄πππ πππππ πΆπ’ππ£π = 0.6554 β 0.2119 = 0.4435
Therefore, the correct answer is π. π. ππππ
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PROBLEM 3: It was found that the mean length of 100 parts produced by a machine was 20.05 mm with a standard deviation of 0.02 mm. The probability that a part selected at random would have a length between 20.06 mm and 20.07 mm is most close to: A. 12.11% B. 14.98% C. 17.30% D. 21.48%
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SOLUTION 3: The formula for CALCULATING A Z-SCORE can be referenced under the topic of NORMAL DISTRIBUTION (GAUSSIAN DISTRIBUTION) on page 39 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. Every normal random variable π₯ can be transformed into a z-score using the equation below:
π=
π₯βπ π
Where: β’ π₯ is the normal random variable β’ π is the mean β’ π is the standard deviation of π₯ To analyze this problem, we will calculate the z-score and then use the normal distributions tables, to find probabilities associated with the z-scores. We are given the following values to solve for z-scores of each wage range: β’ x] = 20.06 β’ π₯^ = 20.07 β’ π = 20.05 β’ π = 0.02
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For the first group of with a mean length of 20.06 mm, we will calculate the z-score, π] , of the probability a wage falls in that range as:
π§] =
(20.06 β 20.05) = 0.5 0.02
For the second group of with a mean length of 20.07 mm, we will calculate the z-score, π^ , of the probability a wage falls in that range as:
π§^ =
(20.07 β 20.05) =1 0.02
Which means that: π 20.06 < π₯ < 20.07 = π(0.5 < π§ < 1) We then look at the normal distribution tables to find the corresponding probability values for each z-score. A unit normal distribution table is included at the end of the reference handbook section on probability and statistics. In the table, the following notations are utilized: β’ πΉ π₯ = π‘βπ ππππ π’ππππ π‘βπ ππ’ππ£π ππππ β β π‘π π₯ β’ π π₯ = π‘βπ ππππ π’ππππ π‘βπ ππ’ππ£π ππππ π₯ π‘π β β’ π π₯ = π‘βπ ππππ π’ππππ π‘βπ ππ’ππ£π ππππ β π₯ π‘π π₯ β’ πΉ βπ₯ = 1 β πΉ(π₯)
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The NORMAL DISTRIBUTION TABLE can be referenced under the topic of ENGINEERING PROBABILITY AND STATISTICS on page 9.4 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. For the first group of with a mean length of 20.06 ππ, we use the z-score, π] = 0.5, to find the probability the wages are GREATER THAN 20.06 ππ. We use the π(π₯) column to indicate π π₯ > 20.06 and π₯ = 0.5 row, as this is the closet value we can use without interpolating. We use the πΉ(π₯) column as we are looking to calculate the area under the curve from π₯ π‘π β, π€βπππ π₯ = 0.5.
π π₯ > 20.06 = π(0.5) = 0.3829
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For the second group of with a mean length of 20.07 mm, we use the z-score, π^ = 1, to find the probability the wages are LESS THAN 20.07 mm. We use the π(π₯) column to indicate π π₯ < 20.07 and π₯ = 1 row. We use the W(x) column as we are looking to calculate the area under the curve from ββ π‘π π₯, π€βπππ π₯ = 20.07.
π π₯ > 20.07 = π 1 = 0.6827
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We then sum the probabilities of each range to calculate the probability that a part selected at random would have a length between 20.06 mm and 20.07 mm. As we are accounting for the under the curve a factor of 2, we need to divide each of the probabilities by a factor of 2: π π₯ < 20.07 π π₯ > 20.06 0.6827 0.3829 β = β = 0.1498 = 14.98% 2 2 2 2
Therefore, the correct answer is π. ππ. ππ%
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PROBLEM 4: The average life of a certain type of engine is 10 years, with a standard deviation of 2 years. If the manufacturer is willing to replace only 3% of the motors that fail, how long should the guarantee be? Assume the motors follow a normal distribution. A. 5.70 π¦ππππ B. 6.20 π¦ππππ C. 6.75 π¦ππππ D. 7.25 π¦ππππ
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SOLUTION 4: The FORMULA FOR CALCULATING A Z-SCORE can be referenced under the topic of NORMAL DISTRIBUTION (GAUSSIAN DISTRIBUTION) on page 39 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. In this problem we need to work backwards to determine the value (in years) that will give the bottom 3% of the distribution. These are the motors that the manufacturer is willing to replace under the guarantee. To analyze this problem, knowing it is normally distribution, we will use the normal distribution able to find the area under the normal distribution curve. We are concerned with the area under the normal distribution curve that is reflected by the 3% that the manufacturer willing to replace.
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As we know the area under the curve, we can work backwards to use the normal distribution table to determine the z-score associated with the probability of 97%. We use the πΉ(π₯) column as we are looking to calculate the area under the curve from π₯ π‘π β, π€βπππ π₯ = 0.9713, as this is the closest value to 0.97 or 97% in the πΉ(π₯) column.
We find that for π (0.97) the corresponding z-score is π§ = β1.9. Every normal random variable π₯ can be transformed into a z-score using the equation below:
π=
π₯βπ π
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Where: β’ π₯ is the normal random variable β’ π is the mean β’ π is the standard deviation of π₯ To analyze this problem, we will calculate the z-score and then use the normal distributions tables, to find probabilities associated with the z-score. We are given the following values to solve for the z-score: β’ x] = πΏπππ ππ πππ‘ππ β’ π₯^ = πΊπ’πππππ‘ππ ππππππ β’ π = 10 β’ π=2 Plugging in the known values and calculated z-score, we can solve for the number of years for the guarantee period:
β1.9 =
π₯ β 10 2
Solving for π₯, we find the guarantee period should be 6.20 π¦ππππ .
The correct answer is π. π. ππ π²πππ«π¬
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