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Service Times In An Automobile Repair Shop Tend To Follow Which Probability Distribution?

Written By Wednesday, April 6, 2022 Add Comment Edit

Learning Outcomes

  • Recognize the uniform probability distribution and employ it appropriately

The uniform distribution is a continuous probability distribution and is concerned with events that are as probable to occur. When working out problems that take a uniform distribution, exist careful to note if the information is inclusive or exclusive.

Example

The data in the table below are 55 smiling times, in seconds, of an eight-week-old baby.

10.4 xix.6 18.8 13.9 17.viii 16.8 21.6 17.nine 12.5 11.one 4.9
12.8 fourteen.8 22.8 20.0 15.ix 16.3 13.four 17.ane 14.5 19.0 22.viii
1.iii 0.seven eight.9 xi.nine ten.9 vii.3 5.ix iii.7 17.9 xix.2 9.eight
5.8 6.9 2.vi 5.8 21.vii eleven.eight 3.4 two.i 4.five vi.3 10.vii
8.nine ix.4 9.4 7.6 10.0 3.3 vi.7 vii.eight 11.6 13.8 18.six

The sample mean[latex]=11.49[/latex] and the sample standard deviation[latex]=six.23[/latex].

We will assume that the grinning times, in seconds, follow a uniform distribution between cipher and 23 seconds, inclusive. This means that any smiling time from zip to and including 23 seconds is equally likely. The histogram that could be constructed from the sample is an empirical distribution that closely matches the theoretical uniform distribution.

Allow [latex]X=[/latex] length, in seconds, of an eight-calendar week-old babe's grinning.

The notation for the uniform distribution is [latex]10{\sim}U(a,b)[/latex] where [latex]a=[/latex] the lowest value of 10 and [latex]b=[/latex] the highest value of x.

The probability density function is [latex]{f{{({x})}}}=\frac{{ane}}{{{b}-{a}}}[/latex] for [latex]a{\leq}10{\leq}b[/latex].

For this example, [latex]Ten{\sim}U(0,23)[/latex] and [latex]{f{{({x})}}}=\frac{{1}}{{{23}-{0}}}[/latex] for [latex]0{\leq}X{\leq}23[/latex].

Formulas for the theoretical hateful and standard divergence are [latex]{\mu}=\frac{{{a}+{b}}}{{two}}{\quad\text{and}\quad}{\sigma}=\sqrt{{\frac{{{({b}-{a})}^{{2}}}}{{12}}}}[/latex]

For this problem, the theoretical mean and standard deviation are [latex]{\mu}=\frac{{{0}+{23}}}{{2}}={11.50} \text{ seconds}{\quad\text{and}\quad}{\sigma}=\sqrt{{\frac{{{({23}-{0})}^{{2}}}}{{12}}}}={6.64} \text{ seconds}[/latex]

Notice that the theoretical mean and standard difference are shut to the sample mean and standard divergence in this case.

Try It

The data that follow are the number of passengers on 35 different lease fishing boats. The sample mean [latex]=7.9[/latex] and the sample standard deviation [latex]=4.33[/latex]. The data follow a uniform distribution where all values between and including zippo and xiv are equally likely. State the values ofa and b. Write the distribution in proper notation, and calculate the theoretical mean and standard difference.

1 12 4 10 4 14 eleven
7 11 iv 13 ii four six
3 x 0 12 vi ix 10
5 thirteen 4 10 xiv 12 11
vi 10 11 0 xi 13 2

a is goose egg; b is 14; [latex]X{\sim}U(0,14)[/latex]; [latex]\mu=7[/latex] passengers; [latex]\sigma=4.04[/latex] passengers

Example

  1. Refer to the previous example. What is the probability that a randomly chosen eight-calendar week-quondam baby smiles between two and 18 seconds?
  2. Observe the 90th percentile for an eight-week-erstwhile infant's smile time.
  3. Discover the probability that a random 8-week-former baby smiles more than 12 seconds knowing that the baby smiles more than 8 seconds.

Try It

A distribution is given as [latex]X{\sim}U(0,20)[/latex]. What is [latex]P(2<ten<18)[/latex]? Find the 90th percentile.

[latex]P(2<x<18)=0.8[/latex]; 90th percentile[latex]=18[/latex]

Example

The amount of time, in minutes, that a person must wait for a bus is uniformly distributed between zero and 15 minutes, inclusive.

  1. What is the probability that a person waits fewer than 12.v minutes?
  2. On the boilerplate, how long must a person wait? Find the mean, μ, and the standard deviation, σ.
  3. Xc percent of the fourth dimension, the time a person must await falls below what value? This asks for the 90th percentile.

Solution

  1. Let X = the number of minutes a person must wait for a bus. a = 0 and b = 15. Ten~ U(0, 15). Write the probability density office. [latex]{f{{({x})}}}=\frac{{1}}{{{15}-{0}}}=\frac{{1}}{{15}}[/latex] for 0 ≤x ≤ 15.FindP (x < 12.five). Depict a graph.
    [latex]{P}{({x}{<}{thou})}={(\text{base})}{(\text{acme})}={({12.v}-{0})}{(\frac{{1}}{{15}})}={0.8333}[/latex]
    The probability a person waits less than 12.5 minutes is 0.8333.
    This shows the graph of the function f(x) = 1/15. A horiztonal line ranges from the point (0, 1/15) to the point (15, 1/15). A vertical line extends from the x-axis to the end of the line at point (15, 1/15) creating a rectangle. A region is shaded inside the rectangle from x = 0 to x = 12.5.
  2. [latex]{\mu}=\frac{{{a}+{b}}}{{2}}=\frac{{{15}+{0}}}{{two}}={vii.5}[/latex]. On the average, a person must expect 7.5 minutes.[latex]{\sigma}=\sqrt{{\frac{{{({b}-{a})}^{{ii}}}}{{12}}}}=\sqrt{{\frac{{{({15}-{0})}^{{ii}}}}{{12}}}}={4.iii}[/latex]The standard deviation is four.three minutes.
  3. Detect the 90th percentile. Draw a graph. Let thou = the 90th percentile.
    [latex]{P}(10<k)=(\text{base of operations})(\text{summit})=(k-0)(\frac{ane}{fifteen})[/latex]
    [latex]{0.90}=(k)(\frac{1}{15})[/latex]
    [latex]g=(0.ninety)(xv)=13.5[/latex]
    k is sometimes called a critical value.The 90th percentile is 13.five minutes. Xc pct of the time, a person must expect at nigh 13.5 minutes.
    f(X)=1/15 graph displaying a boxed region consisting of a horizontal line extending to the right from point 1/15 on the y-axis, a vertical upward line from an arbitrary point on the x-axis, and the x and y-axes. A shaded region from points 0-k occurs within this area. The area of this probability region is equal to 0.90.

Try It

The total duration of baseball game games in the major league in the 2011 season is uniformly distributed between 447 hours and 521 hours inclusive.

  1. Find a and b and describe what they stand for.
  2. Write the distribution.
  3. Find the mean and the standard deviation.
  4. What is the probability that the elapsing of games for a team for the 2011 season is between 480 and 500 hours?
  5. What is the 65th percentile for the duration of games for a team for the 2011 season?
  1. a is 447, and b is 521. a is the minimum duration of games for a team for the 2011 flavor, and b is the maximum duration of games for a team for the 2011 season.
  2. 10 ~ U (447, 521).
  3. μ = 484, and σ = 21.36
  4. P(480 < x < 500) = 0.2703
  5. 65th percentile is 495.1 hours.

Example 4

Suppose the time it takes a ix-year former to eat a donut is between 0.5 and 4 minutes, inclusive. LetTen = the time, in minutes, information technology takes a nine-year former child to eat a donut. So X~ U (0.5, 4).

  1. The probability that a randomly selected nine-year old child eats a donut in at least two minutes is _______.
  2. Find the probability that a different nine-year old kid eats a donut in more than than ii minutes given that the child has already been eating the donut for more than 1.v minutes.

Solution

  1. 0.5714
  2. This question has a conditional probability. You are asked to find the probability that a nine-yr old child eats a donut in more than than ii minutes given that the child has already been eating the donut for more than 1.5 minutes. Solve the trouble ii dissimilar means (run across Example 3). You must reduce the sample space.
    1. Commencement way: Since you know the child has already been eating the donut for more than than ane.5 minutes, you are no longer starting at a = 0.5 minutes. Your starting point is 1.5 minutes. Write a newf(ten):[latex]{f{{({x})}}}=\frac{{1}}{{{4}-{ane.5}}}={25} \text{ for } {1.5}\leq{x}\leq{4}[/latex]FindP(x > 2|10 > one.5). Describe a graph.
      f(X)=2/5 graph displaying a boxed region consisting of a horizontal line extending to the right from point 2/5 on the y-axis, a vertical upward line from points 1.5 and 4 on the x-axis, and the x-axis. A shaded region from points 2-4 occurs within this area.[latex]{P}{({x}{>}{2}|{10}{>}{one.5})}={(\text{base})}{(\text{new top})}={({4}-{two})}{(\frac{{2}}{{5}})}=\frac{{4}}{{5}}[/latex]
      The probability that a nine-year former child eats a donut in more than two minutes given that the child has already been eating the donut for more than i.5 minutes is [latex]\frac{{4}}{{5}}[/latex].
    2. 2d way: Describe the original graph for X ~ U (0.5, 4). Apply the conditional formula[latex]{P}{({ten}{>}{2}{mid}{x}{>}{1.v})}=\frac{{{P}{({x}{>}{2} \text{ AND } {ten}{>}{1.5})}}}{{{P}{({x}{>}{1.five})}}}=\frac{{{P}{({ten}{>}{2})}}}{{{P}{({x}{>}{i.5})}}}=\frac{{\frac{{2}}{{3.5}}}}{{\frac{{2.5}}{{iii.five}}}}={0.8}=\frac{{four}}{{5}}[/latex]

Try Information technology

Suppose the time it takes a educatee to finish a quiz is uniformly distributed between 6 and 15 minutes, inclusive. Let10 = the time, in minutes, it takes a pupil to finish a quiz. Then X ~ U (6, 15).

Find the probability that a randomly selected student needs at least 8 minutes to complete the quiz. Then find the probability that a different student needs at least eight minutes to finish the quiz given that she has already taken more than 7 minutes.

P (10 > 8) = 0.7778

P (x > 8 | x > 7) = 0.875

Case 5

Ace Heating and Air Workout Service finds that the corporeality of time a repairman needs to fix a furnace is uniformly distributed between 1.v and four hours. Let10 = the time needed to fix a furnace. Then ten ~ U (i.5, 4).

  1. Find the probability that a randomly selected furnace repair requires more two hours.
  2. Detect the probability that a randomly selected furnace repair requires less than three hours.
  3. Find the 30th percentile of furnace repair times.
  4. The longest 25% of furnace repair times take at least how long? (In other words: observe the minimum time for the longest 25% of repair times.) What percentile does this stand for?
  5. Find the hateful and standard deviation

Solution

  1. To noticef(x): [latex]{f{{({x})}}}=\frac{{1}}{{{iv}-{1.5}}}={12.5} \text{ so } {f{{({x})}}}={0.4}[/latex]
    P(10 > 2) = (base)(height) = (4 – 2)(0.4) = 0.8
    This shows the graph of the function f(x) = 0.4. A horiztonal line ranges from the point (1.5, 0.4) to the point (4, 0.4). Vertical lines extend from the x-axis to the graph at x = 1.5 and x = 4 creating a rectangle. A region is shaded inside the rectangle from x = 2 to x = 4.
    Uniform Distribution between 1.5 and 4 with shaded expanse between two and four representing the probability that the repair fourth dimension
    10 is greater than two
  2. P(10 < 3) = (base)(height) = (3 – one.v)(0.4) = 0.6The graph of the rectangle showing the entire distribution would remain the same. Even so the graph should be shaded betwixt
    10 = 1.5 and x = iii. Note that the shaded expanse starts at x = 1.5 rather than at x = 0; since Ten ~ U (1.5, iv), ten can non be less than 1.5.
    This shows the graph of the function f(x) = 0.4. A horiztonal line ranges from the point (1.5, 0.4) to the point (4, 0.4). Vertical lines extend from the x-axis to the graph at x = 1.5 and x = 4 creating a rectangle. A region is shaded inside the rectangle from x = 1.5 to x = 3.Compatible Distribution between i.5 and four with shaded expanse betwixt 1.v and iii representing the probability that the repair time10 is less than 3
  3. This shows the graph of the function f(ten) = 0.4. A horiztonal line ranges from the point (1.five, 0.4) to the indicate (4, 0.four). Vertical lines extend from the 10-axis to the graph at x = 1.5 and ten = 4 creating a rectangle. A region is shaded within the rectangle from ten = i.5 to x = k. The shaded expanse represents P(10 < k) = 0.3.
    Uniform Distribution between 1.five and 4 with an area of 0.thirty shaded to the left, representing the shortest 30% of repair times.
    P (10 < g) = 0.30
    P(x < thousand) = (base)(height) = (thousand – 1.v)(0.4)0.iii = (k – 1.5) (0.four); Solve to find one thousand:0.75 =chiliad – ane.v, obtained past dividing both sides by 0.four
    k = 2.25 , obtained by adding 1.v to both sidesThe 30th percentile of repair times is 2.25 hours. 30% of repair times are 2.5 hours or less.

  4. Uniform Distribution between 1.5 and 4 with an area of 0.25 shaded to the right representing the longest 25% of repair times.P(ten > k) = 0.25
    P(10 > m) = (base)(height) = (iv – thousand)(0.4)0.25 = (4 –grand)(0.four); Solve for 1000:0.625 = 4 −k, obtained past dividing both sides by 0.iv

−3.375 = −chiliad, obtained by subtracting four from both sides: k = 3.375

The longest 25% of furnace repairs take at least iii.375 hours (3.375 hours or longer).

Note: Since 25% of repair times are iii.375 hours or longer, that means that 75% of repair times are iii.375 hours or less. three.375 hours is the 75th percentile of furnace repair times.

  • [latex]{\mu}={\frac{a+b}{2}}\text{ and }{\sigma}=\sqrt{\frac{(b-a)^2}{12}}[/latex]
    [latex]{\mu}=\frac{one.v+4}{2}=ii.75\text{ hours and }{\sigma}=\sqrt{\frac{(4-one.v)^2}{12}}= 0.7217 \text{ hours}[/latex]

Try Information technology

The amount of time a service technician needs to change the oil in a automobile is uniformly distributed betwixt 11 and 21 minutes. Let10 = the time needed to change the oil on a car.

  1. Write the random variable X in words. X = __________________.
  2. Write the distribution.
  3. Graph the distribution.
  4. Detect P (x > 19).
  5. Find the 50th percentile.
  1. Permit 10 = the time needed to alter the oil in a car.
  2. X ~ U (11, 21).
  3. This graph shows a uniform distribution. The horizontal axis ranges from 405 to 525. The distribution is modeled by a rectangle extending from x = 447 to x = 521.
  4. P (x > nineteen) = 0.2
  5. the 50th percentile is 16 minutes.

References

McDougall, John A. The McDougall Program for Maximum Weight Loss. Plume, 1995.

Concept Review

If10 has a uniform distribution where a < 10 < b or a ≤ 10 ≤ b, then X takes on values betwixt a and b (may include a and b). All values x are equally likely. Nosotros write 10 ∼ U(a, b). The mean of Ten is [latex]{\mu}=\frac{{{a}+{b}}}{{2}}[/latex]. X is continuous.

The graph shows a rectangle with total area equal to 1. The rectangle extends from x = a to x = b on the x-axis and has a height of 1/(b-a).

The probabilityP(c < 10 < d) may exist found by computing the area under f(ten), between c and d. Since the corresponding area is a rectangle, the surface area may exist found just past multiplying the width and the height.

Formula Review

X = a real number between a and b (in some instances, X can have on the values a and b). a = smallest X; b = largest X

Ten ~ U (a, b)

The mean is [latex]\mu=\frac{{{a}+{b}}}{{2}}[/latex]

The standard departure is [latex]\sigma=\sqrt{{\frac{{({b}-{a})}^{{2}}}{{12}}}}[/latex]

Probability density role: [latex]{f{{({ten})}}}=\frac{{1}}{{{b}-{a}}} \text{ for } {a}\leq{X}\leq{b}[/latex]

Area to the Left ofx: [latex]{P}{({X}{<}{x})}={({x}-{a})}{(\frac{{1}}{{{b}-{a}}})}[/latex]

Expanse to the Correct ofx: [latex]{P}{({X}{>}{x})}={({b}-{x})}{(\frac{{1}}{{{b}-{a}}})}[/latex]

Expanse Betweenc and d: [latex]{P}{({c}{<}{x}{<}{d})}={(\text{base})}{(\text{height})}={({d}-{c})}{(\frac{{ane}}{{{b}-{a}}})}[/latex]

Uniform:X ~ U(a, b) where a < x < b

  • pdf: [latex]{f{{({10})}}}=\frac{{i}}{{{b}-{a}}}[/latex] for a ≤ 10 ≤ b
  • cdf: P(X ≤ ten) = [latex]\frac{{{x}-{a}}}{{{b}-{a}}}[/latex]
  • mean: [latex]\mu=\frac{{{a}+{b}}}{{2}}[/latex]
  • standard divergence: [latex]\sigma=\sqrt{{\frac{{({b}-{a})}^{{2}}}{{12}}}}[/latex]
  • P(c < X < d) = (d – c)

Source: https://courses.lumenlearning.com/introstats1/chapter/the-uniform-distribution/

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