{"id":229,"date":"2022-05-18T16:37:46","date_gmt":"2022-05-18T16:37:46","guid":{"rendered":"https:\/\/pressbooks.ccconline.org\/accintrostats\/chapter\/the-uniform-distribution\/"},"modified":"2022-08-10T19:45:10","modified_gmt":"2022-08-10T19:45:10","slug":"the-uniform-distribution","status":"publish","type":"chapter","link":"https:\/\/pressbooks.ccconline.org\/accintrostats\/chapter\/the-uniform-distribution\/","title":{"raw":"Chapter 6.3: The Uniform Distribution","rendered":"Chapter 6.3: The Uniform Distribution"},"content":{"raw":" \r\n

The uniform distribution is a continuous probability distribution and is concerned with events that are equally likely to occur. When working out problems that have a uniform distribution, be careful to note if the data is inclusive or exclusive of endpoints.<\/p>\r\n\r\n

\r\n\r\nThe data in (Figure)<\/a> are 55 smiling times, in seconds, of an eight-week-old baby.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
10.4<\/td>\r\n19.6<\/td>\r\n18.8<\/td>\r\n13.9<\/td>\r\n17.8<\/td>\r\n16.8<\/td>\r\n21.6<\/td>\r\n17.9<\/td>\r\n12.5<\/td>\r\n11.1<\/td>\r\n4.9<\/td>\r\n<\/tr>\r\n
12.8<\/td>\r\n14.8<\/td>\r\n22.8<\/td>\r\n20.0<\/td>\r\n15.9<\/td>\r\n16.3<\/td>\r\n13.4<\/td>\r\n17.1<\/td>\r\n14.5<\/td>\r\n19.0<\/td>\r\n22.8<\/td>\r\n<\/tr>\r\n
1.3<\/td>\r\n0.7<\/td>\r\n8.9<\/td>\r\n11.9<\/td>\r\n10.9<\/td>\r\n7.3<\/td>\r\n5.9<\/td>\r\n3.7<\/td>\r\n17.9<\/td>\r\n19.2<\/td>\r\n9.8<\/td>\r\n<\/tr>\r\n
5.8<\/td>\r\n6.9<\/td>\r\n2.6<\/td>\r\n5.8<\/td>\r\n21.7<\/td>\r\n11.8<\/td>\r\n3.4<\/td>\r\n2.1<\/td>\r\n4.5<\/td>\r\n6.3<\/td>\r\n10.7<\/td>\r\n<\/tr>\r\n
8.9<\/td>\r\n9.4<\/td>\r\n9.4<\/td>\r\n7.6<\/td>\r\n10.0<\/td>\r\n3.3<\/td>\r\n6.7<\/td>\r\n7.8<\/td>\r\n11.6<\/td>\r\n13.8<\/td>\r\n18.6<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nThe sample mean = 11.49 and the sample standard deviation = 6.23.\r\n

We will assume that the smiling times, in seconds, follow a uniform distribution between zero and 23 seconds, inclusive. This means that any smiling time from zero to and including 23 seconds is equally likely<\/span>. The histogram that could be constructed from the sample is an empirical distribution that closely matches the theoretical uniform distribution.<\/p>\r\nLet X<\/em> = length, in seconds, of an eight-week-old baby's smile.\r\n\r\nThe notation for the uniform distribution is\r\n\r\nX<\/em> ~ U<\/em>(a<\/em>, b<\/em>) where a<\/em> = the lowest value of x<\/em> and b<\/em> = the highest value of x<\/em>.\r\n\r\nThe probability density function is f<\/em>(x<\/em>) = \\(\\frac{1}{b-a}\\) for a<\/em> \u2264 x<\/em> \u2264 b<\/em>.\r\n\r\nFor this example, X<\/em> ~ U<\/em>(0, 23) and f<\/em>(x<\/em>) = \\(\\frac{1}{23-0}\\) for 0 \u2264 X<\/em> \u2264 23.\r\n

Formulas for the theoretical mean and standard deviation are<\/p>\r\n\\(\\mu =\\frac{a+b}{2}\\) and \\(\\sigma =\\sqrt{\\frac{{\\left(b-a\\right)}^{2}}{12}}\\)\r\n

For this problem, the theoretical mean and standard deviation are<\/p>\r\n\u03bc<\/em> = \\(\\frac{0\\text{\u00a0}+\\text{\u00a0}23}{2}\\) = 11.50 seconds and \u03c3<\/em> = \\(\\sqrt{\\frac{{\\left(23\\text{\u00a0}-\\text{\u00a0}0\\right)}^{2}}{12}}\\) = 6.64 seconds.\r\n\r\nNotice that the theoretical mean and standard deviation are close to the sample mean and standard deviation in this example.\r\n\r\n<\/div>\r\n

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Try It<\/div>\r\n
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The data that follow are the number of passengers on 35 different charter fishing boats. The sample mean = 7.9 and the sample standard deviation = 4.33. The data follow a uniform distribution where all values between and including zero and 14 are equally likely. State the values of a<\/em> and b<\/em>. Write the distribution in proper notation, and calculate the theoretical mean and standard deviation.<\/p>\r\n\r\n<\/colgroup>\r\n\r\n\r\n\r\n\r\n\r\n\r\n
1<\/td>\r\n12<\/td>\r\n4<\/td>\r\n10<\/td>\r\n4<\/td>\r\n14<\/td>\r\n11<\/td>\r\n<\/tr>\r\n
7<\/td>\r\n11<\/td>\r\n4<\/td>\r\n13<\/td>\r\n2<\/td>\r\n4<\/td>\r\n6<\/td>\r\n<\/tr>\r\n
3<\/td>\r\n10<\/td>\r\n0<\/td>\r\n12<\/td>\r\n6<\/td>\r\n9<\/td>\r\n10<\/td>\r\n<\/tr>\r\n
5<\/td>\r\n13<\/td>\r\n4<\/td>\r\n10<\/td>\r\n14<\/td>\r\n12<\/td>\r\n11<\/td>\r\n<\/tr>\r\n
6<\/td>\r\n10<\/td>\r\n11<\/td>\r\n0<\/td>\r\n11<\/td>\r\n13<\/td>\r\n2<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n
\r\n
\r\n
\r\n\r\na. Refer to (Figure)<\/a>. What is the probability that a randomly chosen eight-week-old baby smiles between two and 18 seconds?\r\n\r\n<\/div>\r\n
\r\n

P<\/em>(2 < x<\/em> < 18) = (base)(height) = (18 \u2013 2)\\(\\left(\\frac{1}{23}\\right)\\) = \\(\\frac{16}{23}\\).<\/p>\r\n\r\n

\"This<\/span><\/div>\r\n<\/div>\r\n<\/div>\r\n
\r\n
\r\n\r\nb. Find the 90th<\/sup> percentile for an eight-week-old baby's smiling time.\r\n\r\n<\/div>\r\n
\r\n\r\nb. Ninety percent of the smiling times fall below the 90th<\/sup> percentile, k<\/em>, so P<\/em>(x<\/em> < k<\/em>) = 0.90.\r\n\r\n\\(P\\left(x<k\\right)=0.90\\)\r\n

\\(\\left(\\text{base}\\right)\\left(\\text{height}\\right)=0.90\\)<\/p>\r\n\\(\\text{(}k-0\\text{)}\\left(\\frac{1}{23}\\right)=0.90\\)\r\n\r\n\\(k=\\left(23\\right)\\left(0.90\\right)=20.7\\)\r\n\r\n[caption id=\"\" align=\"alignnone\" width=\"487\"]\"Shaded 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 = k. The shaded area represents P(x < k)[\/caption]\r\n\r\n<\/div>\r\n<\/div>\r\n

\r\n
\r\n\r\nc. Find the probability that a random eight-week-old baby smiles more than 12 seconds KNOWING<\/strong> that the baby smiles MORE THAN EIGHT SECONDS<\/strong>.\r\n\r\n<\/div>\r\n
\r\n

c. This probability question is a conditional<\/strong>. You are asked to find the probability that an eight-week-old baby smiles more than 12 seconds when you already know<\/strong> the baby has smiled for more than eight seconds.<\/p>\r\n

Find P<\/em>(x<\/em> > 12|x<\/em> > 8) There are two ways to do the problem. For the first way<\/strong>, use the fact that this is a conditional<\/strong> and changes the sample space. The graph illustrates the new sample space. You already know the baby smiled more than eight seconds.<\/p>\r\n

Write a new<\/strong>f<\/em>(x<\/em>): f<\/em>(x<\/em>) = \\(\\frac{1}{23\\text{\u00a0}-\\text{\u00a08}}\\) = \\(\\frac{1}{15}\\) for 8 < x<\/em> < 23<\/p>\r\nP<\/em>(x<\/em> > 12|x<\/em> > 8) = (23 \u2212 12)\\(\\left(\\frac{1}{15}\\right)\\) = \\(\\frac{11}{15}\\)\r\n

\"f(X)=1\/15<\/span><\/div>\r\nFor the second way<\/strong>, use the conditional formula from Probability Topics<\/a> with the original distribution X<\/em> ~ U<\/em> (0, 23):\r\n\r\nP<\/em>(A<\/em>|B<\/em>) = \\(\\frac{P\\left(A\\text{\u00a0AND\u00a0}B\\right)}{P\\left(B\\right)}\\)\r\n

For this problem, A<\/em> is (x<\/em> > 12) and B<\/em> is (x<\/em> > 8).<\/p>\r\nSo, P<\/em>(x<\/em> > 12<\/em>|x<\/em> > 8) = \\(\\frac{\\left(x>12\\text{\u00a0AND\u00a0}x>8\\right)}{P\\left(x>8\\right)}=\\frac{P\\left(x>12\\right)}{P\\left(x>8\\right)}=\\frac{\\frac{11}{23}}{\\frac{15}{23}}=\\frac{11}{15}\\)\r\n

\"This<\/span><\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n
\r\n
Try It<\/div>\r\n
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\r\n

A distribution is given as X<\/em> ~ U<\/em> (0, 20). What is P<\/em>(2 < x<\/em> < 18)? Find the 90th<\/sup> percentile.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

\r\n\r\nThe amount of time, in minutes, that a person must wait for a bus is uniformly distributed between zero and 15 minutes, inclusive.\r\n<\/span>\r\n
\r\n
\r\n

a. What is the probability that a person waits fewer than 12.5 minutes?<\/p>\r\n\r\n<\/div>\r\n

\r\n\r\na. Let X<\/em> = the number of minutes a person must wait for a bus. a<\/em> = 0 and b<\/em> = 15. X<\/em> ~ U<\/em>(0, 15). Write the probability density function. f<\/em> (x<\/em>) = \\(\\frac{1}{15\\text{\u00a0}-\\text{\u00a0}0}\\) = \\(\\frac{1}{15}\\) for 0 \u2264 x<\/em> \u2264 15.\r\n\r\nFind P<\/em> (x<\/em> < 12.5). Draw a graph.\r\n\r\n\\(P\\left(x<k\\right)=\\left(\\text{base}\\right)\\left(\\text{height}\\right)=\\left(12.5-0\\right)\\left(\\frac{1}{15}\\right)=0.8333\\)\r\n

The probability a person waits less than 12.5 minutes is 0.8333.<\/p>\r\n\r\n

\"This<\/span><\/div>\r\n<\/div>\r\n<\/div>\r\n
\r\n
\r\n

b. On the average, how long must a person wait? Find the mean, \u03bc<\/em>, and the standard deviation, \u03c3<\/em>.<\/p>\r\n\r\n<\/div>\r\n

\r\n

b. \u03bc<\/em> = \\(\\frac{a\\text{\u00a0}+\\text{\u00a0}b}{2}\\) = \\(\\frac{15\\text{\u00a0}+\\text{\u00a0}0}{2}\\) = 7.5. On the average, a person must wait 7.5 minutes. \r\n<\/span> \r\n<\/span> \u03c3<\/em> = \\(\\sqrt{\\frac{\\left(b-a{\\right)}^{2}}{12}}=\\sqrt{\\frac{\\left(\\mathrm{15}-0{\\right)}^{2}}{12}}\\) = 4.3. The Standard deviation is 4.3 minutes. \r\n<\/span><\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

\r\n
\r\n\r\nc. Ninety percent of the time, the time a person must wait falls below what value?\r\n
This asks for the 90th<\/sup> percentile.<\/div>\r\n<\/div>\r\n
\r\n\r\nc. Find the 90th<\/sup> percentile. Draw a graph. Let k<\/em> = the 90th<\/sup> percentile. \r\n<\/span> \r\n<\/span>\\(P\\left(x<k\\right)=\\left(\\text{base}\\right)\\left(\\text{height}\\right)=\\left(k-0\\right)\\left(\\frac{1}{15}\\right)\\) \r\n<\/span> \r\n<\/span> \\(0.90=\\left(k\\right)\\left(\\frac{1}{15}\\right)\\) \r\n<\/span> \r\n<\/span>\\(k=\\left(0.90\\right)\\left(15\\right)=13.5\\) \r\n<\/span> \r\n<\/span> k<\/em> is sometimes called a critical value. \r\n<\/span> \r\n<\/span>The 90th<\/sup> percentile is 13.5 minutes. Ninety percent of the time, a person must wait at most 13.5 minutes.\r\n
\"f(X)=1\/15<\/span><\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n
\r\n
Try It<\/div>\r\n
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\r\n

The total duration of baseball games in the major league in the 2011 season is uniformly distributed between 447 hours and 521 hours inclusive.<\/p>\r\n\r\n

    \r\n \t
  1. Find a<\/em> and b<\/em> and describe what they represent.<\/li>\r\n \t
  2. Write the distribution.<\/li>\r\n \t
  3. Find the mean and the standard deviation.<\/li>\r\n \t
  4. What is the probability that the duration of games for a team for the 2011 season is between 480 and 500 hours?<\/li>\r\n \t
  5. What is the 65th<\/sup> percentile for the duration of games for a team for the 2011 season?<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n
    \r\n

    Suppose the time it takes a nine-year old to eat a donut is between 0.5 and 4 minutes, inclusive. Let X<\/em> = the time, in minutes, it takes a nine-year old child to eat a donut. Then X<\/em> ~ U<\/em> (0.5, 4).\r\n<\/span><\/p>\r\n\r\n

    \r\n
    \r\n

    a. The probability that a randomly selected nine-year old child eats a donut in at least two minutes is _______.<\/p>\r\n\r\n<\/div>\r\n

    \r\n

    a. 0.5714\r\n<\/span><\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

    \r\n
    \r\n\r\nb. Find the probability that a different nine-year old child eats a donut in more than two minutes given that the child has already been eating the donut for more than 1.5 minutes.\r\n

    The second question has a conditional probability<\/span>. You are asked to find the probability that a nine-year old child eats a donut in more than two minutes given that the child has already been eating the donut for more than 1.5 minutes. Solve the problem two different ways (see (Figure)<\/a>). You must reduce the sample space. First way<\/strong>: Since you know the child has already been eating the donut for more than 1.5 minutes, you are no longer starting at a<\/em> = 0.5 minutes. Your starting point is 1.5 minutes.<\/p>\r\n

    Write a new f<\/em>(x<\/em>):<\/strong><\/p>\r\n

    f<\/em>(x<\/em>) = \\(\\frac{1}{4-1.5}\\) = \\(\\frac{2}{5}\\) for 1.5 \u2264 x<\/em> \u2264 4.<\/p>\r\n

    Find P<\/em>(x<\/em> > 2|x<\/em> > 1.5). Draw a graph.<\/p>\r\n\r\n

    \"f(X)=2\/5<\/span><\/div>\r\n

    P<\/em>(x<\/em> > 2<\/em>|x<\/em> > 1.5) = (base)(new height) = (4 \u2212 2)\\(\\left(\\frac{2}{5}\\right)=\\frac{4}{5}\\)<\/p>\r\n\r\n<\/div>\r\n

    \r\n

    b. \\(\\frac{4}{5}\\)<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

    The probability that a nine-year old child eats a donut in more than two minutes given that the child has already been eating the donut for more than 1.5 minutes is \\(\\frac{4}{5}\\).<\/p>\r\n

    Second way:<\/strong> Draw the original graph for X<\/em> ~ U<\/em> (0.5, 4). Use the conditional formula<\/p>\r\n

    P<\/em>(x<\/em> > 2|x<\/em> > 1.5) = \\(\u00a0\\frac{P\\left(x>2\\text{\u00a0AND\u00a0}x>1.5\\right)}{P\\left(x>\\text{1}\\text{.5}\\right)}=\\frac{P\\left(x>2\\right)}{P\\left(x>1.5\\right)}=\\frac{\\frac{2}{3.5}}{\\frac{2.5}{3.5}}=\\text{0}\\text{.8}=\\frac{4}{5}\\)<\/p>\r\n\r\n<\/div>\r\n

    \r\n
    Try It<\/div>\r\n
    \r\n
    \r\n

    Suppose the time it takes a student to finish a quiz is uniformly distributed between six and 15 minutes, inclusive. Let X<\/em> = the time, in minutes, it takes a student to finish a quiz. Then X<\/em> ~ U<\/em> (6, 15).<\/p>\r\n

    Find the probability that a randomly selected student needs at least eight 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 seven minutes.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

    \r\n

    Ace Heating and Air Conditioning Service finds that the amount of time a repairman needs to fix a furnace is uniformly distributed between 1.5 and four hours. Let x<\/em> = the time needed to fix a furnace. Then x<\/em> ~ U<\/em> (1.5, 4).<\/p>\r\n\r\n

    \r\n
    \r\n
      \r\n \t
    1. Find the probability that a randomly selected furnace repair requires more than two hours.<\/li>\r\n \t
    2. Find the probability that a randomly selected furnace repair requires less than three hours.<\/li>\r\n \t
    3. Find the 30th<\/sup> percentile of furnace repair times.<\/li>\r\n \t
    4. The longest 25% of furnace repair times take at least how long? (In other words: find the minimum time for the longest 25% of repair times.) What percentile does this represent?<\/li>\r\n \t
    5. Find the mean and standard deviation<\/li>\r\n<\/ol>\r\n<\/div>\r\n
      \r\n

      a. To find f<\/em>(x<\/em>): f<\/em> (x<\/em>) = \\(\\frac{1}{4\\text{\u00a0}-\\text{\u00a0}1.5}\\) = \\(\\frac{1}{2.5}\\) so f<\/em>(x<\/em>) = 0.4<\/p>\r\n

      P<\/em>(x<\/em> > 2) = (base)(height) = (4 \u2013 2)(0.4) = 0.8<\/p>\r\n\r\n

      \r\n
      Uniform Distribution between 1.5 and four with shaded area between two and four representing the probability that the repair time x<\/em> is greater than two<\/div>\r\n\"This<\/span>\r\n\r\n<\/div>\r\n<\/div>\r\n
      \r\n

      b. P<\/em>(x<\/em> < 3) = (base)(height) = (3 \u2013 1.5)(0.4) = 0.6<\/p>\r\n

      The graph of the rectangle showing the entire distribution would remain the same. However the graph should be shaded between x<\/em> = 1.5 and x<\/em> = 3. Note that the shaded area starts at x<\/em> = 1.5 rather than at x<\/em> = 0; since X<\/em> ~ U<\/em> (1.5, 4), x<\/em> can not be less than 1.5.<\/p>\r\n\r\n

      \r\n
      Uniform Distribution between 1.5 and four with shaded area between 1.5 and three representing the probability that the repair time x<\/em> is less than three<\/div>\r\n\"This<\/span>\r\n\r\n<\/div>\r\n<\/div>\r\n
      \r\n

      c.<\/p>\r\n\r\n

      \r\n
      Uniform Distribution between 1.5 and 4 with an area of 0.30 shaded to the left, representing the shortest 30% of repair times.<\/div>\r\n\r\n[caption id=\"\" align=\"alignnone\" width=\"487\"]\"Shaded 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 = k. The shaded area represents P(x < k)[\/caption]\r\n\r\n<\/div>\r\n

      \r\n<\/span>P<\/em> (x<\/em> < k<\/em>) = 0.30 \r\n<\/span> P<\/em>(x<\/em> < k<\/em>) = (base)(height) = (k<\/em> \u2013 1.5)(0.4) \r\n<\/span>0.3 = (k<\/em> \u2013 1.5) (0.4)<\/strong>; Solve to find k<\/em>: \r\n<\/span>0.75 = k<\/em> \u2013 1.5, obtained by dividing both sides by 0.4 \r\n<\/span>k<\/em> = 2.25 <\/strong>, obtained by adding 1.5 to both sides \r\n<\/span>The 30th<\/sup> percentile of repair times is 2.25 hours. 30% of repair times are 2.5 hours or less.<\/p>\r\n\r\n<\/div>\r\n

      \r\n

      d.<\/p>\r\n\r\n

      \r\n
      Uniform Distribution between 1.5 and 4 with an area of 0.25 shaded to the right representing the longest 25% of repair times.<\/div>\r\n\"\"<\/span>\r\n\r\n<\/div>\r\n

      \r\n<\/span>P<\/em>(x<\/em> > k<\/em>) = 0.25 \r\n<\/span> P<\/em>(x<\/em> > k<\/em>) = (base)(height) = (4 \u2013 k<\/em>)(0.4) \r\n<\/span>0.25 = (4 \u2013 k<\/em>)(0.4)<\/strong>; Solve for k<\/em>: \r\n<\/span>0.625 = 4 \u2212 k<\/em>, \r\n<\/span>obtained by dividing both sides by 0.4 \r\n<\/span>\u22123.375 = \u2212k<\/em>, \r\n<\/span>obtained by subtracting four from both sides: k<\/em> = 3.375<\/strong> \r\n<\/span>The longest 25% of furnace repairs take at least 3.375 hours (3.375 hours or longer). \r\n<\/span>Note:<\/strong> Since 25% of repair times are 3.375 hours or longer, that means that 75% of repair times are 3.375 hours or less. 3.375 hours is the 75th<\/sup> percentile<\/strong> of furnace repair times.<\/p>\r\n\r\n<\/div>\r\n

      \r\n

      e. \\(\\mu =\\frac{a+b}{2}\\) and \\(\\sigma =\\sqrt{\\frac{{\\left(b-a\\right)}^{2}}{12}}\\) \r\n<\/span>\\(\\mu =\\frac{1.5+4}{2}=2.75\\) hours and \\(\\sigma =\\sqrt{\\frac{{\\left(4\u20131.5\\right)}^{2}}{12}}=0.7217\\) hours<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

      \r\n
      Try It<\/div>\r\n
      \r\n
      \r\n

      The amount of time a service technician needs to change the oil in a car is uniformly distributed between 11 and 21 minutes. Let X<\/em> = the time needed to change the oil on a car.<\/p>\r\n\r\n

        \r\n \t
      1. Write the random variable X<\/em> in words. X<\/em> = __________________.<\/li>\r\n \t
      2. Write the distribution.<\/li>\r\n \t
      3. Graph the distribution.<\/li>\r\n \t
      4. Find P<\/em> (x<\/em> > 19).<\/li>\r\n \t
      5. Find the 50th<\/sup> percentile.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n
        \r\n

        Chapter Review<\/h3>\r\n

        If X<\/em> has a uniform distribution where a<\/em> < x<\/em> < b<\/em> or a<\/em> \u2264 x<\/em> \u2264 b<\/em>, then X<\/em> takes on values between a<\/em> and b<\/em> (may include a<\/em> and b<\/em>). All values x<\/em> are equally likely. We write X<\/em> \u223c U<\/em>(a<\/em>, b<\/em>). The mean of X<\/em> is \\(\\mu =\\frac{a+b}{2}\\). The standard deviation of X<\/em> is \\(\\sigma =\\sqrt{\\frac{{\\left(b-a\\right)}^{2}}{12}}\\). The probability density function of X<\/em> is \\(f\\left(x\\right)=\\frac{1}{b-a}\\) for a<\/em> \u2264 x<\/em> \u2264 b<\/em>. The cumulative distribution function of X<\/em> is P<\/em>(X<\/em> \u2264 x<\/em>) = \\(\\frac{x-a}{b-a}\\). X<\/em> is continuous.<\/p>\r\n\r\n

        \"The<\/span><\/div>\r\n

        The probability P<\/em>(c<\/em> < X<\/em> < d<\/em>) may be found by computing the area under f<\/em>(x<\/em>), between c<\/em> and d<\/em>. Since the corresponding area is a rectangle, the area may be found simply by multiplying the width and the height.<\/p>\r\n\r\n<\/div>\r\n

        \r\n

        Formula Review<\/h3>\r\nX<\/em> = a real number between a<\/em> and b<\/em> (in some instances, X<\/em> can take on the values a<\/em> and b<\/em>). a<\/em> = smallest X<\/em>; b<\/em> = largest X<\/em>\r\n

        X<\/em> ~ U<\/em> (a, b)<\/p>\r\nThe mean is \\(\\mu =\\frac{a+b}{2}\\)\r\n\r\nThe standard deviation is \\(\\sigma =\\sqrt{\\frac{{\\left(b\\text{\u00a0\u2013\u00a0}a\\right)}^{2}}{12}}\\)\r\n\r\nProbability density function:<\/strong>\\(f\\left(x\\right)=\\frac{1}{b-a}\\) for \\(a\\le X\\le b\\)\r\n\r\nArea to the Left of x<\/em>:<\/strong>P<\/em>(X<\/em> < x<\/em>) = (x<\/em> \u2013 a<\/em>)\\(\\left(\\frac{1}{b-a}\\right)\\)\r\n\r\nArea to the Right of x<\/em>:<\/strong>P<\/em>(X<\/em> > x<\/em>) = (b<\/em> \u2013 x<\/em>)\\(\\left(\\frac{1}{b-a}\\right)\\)\r\n

        Area Between c<\/em> and d<\/em>:<\/strong>P<\/em>(c<\/em> < x<\/em> < d<\/em>) = (base)(height) = (d<\/em> \u2013 c<\/em>)\\(\\left(\\frac{1}{b-a}\\right)\\)<\/p>\r\n

        Uniform: X<\/em> ~ U<\/em>(a<\/em>, b<\/em>) where a<\/em> < x<\/em> < b<\/em><\/p>\r\n\r\n