{"id":53,"date":"2021-01-12T22:19:32","date_gmt":"2021-01-12T22:19:32","guid":{"rendered":"https:\/\/textbooks.jaykesler.net\/introstats\/chapter\/the-uniform-distribution\/"},"modified":"2023-06-26T23:50:51","modified_gmt":"2023-06-26T23:50:51","slug":"the-uniform-distribution","status":"publish","type":"chapter","link":"https:\/\/textbooks.jaykesler.net\/introstats\/chapter\/the-uniform-distribution\/","title":{"rendered":"The Uniform Distribution"},"content":{"raw":"\r\n[latexpage]\r\n<\/span>\r\n
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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

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Example <\/span>5.2<\/span><\/h3>\r\n<\/header>
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The data in Table 5.1<\/a> below are 55 smiling times, in seconds, of an eight-week-old baby.<\/p>\r\n\r\n

\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\n
Table <\/span>5.1<\/span><\/div>\r\n<\/div>\r\n

The sample mean = 11.49 and the sample standard deviation = 6.23.<\/p>\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\n

Let X<\/em> = length, in seconds, of an eight-week-old baby's smile.<\/p>\r\n

The notation for the uniform distribution is<\/p>\r\n

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

The probability density function is $f(x) = \\frac{1}{b-a}$ for a \u2264 x<\/em> \u2264 b<\/em>.<\/em><\/p>\r\n

For this example, X<\/em> ~ U<\/em>(0, 23) and $f(x) = \\frac{1}{23-0}$ or 0 \u2264 X<\/em> \u2264 23.<\/p>\r\n

Formulas for the theoretical mean and standard deviation are<\/p>\r\n$\\mu = \\frac{a+b}{2}$ and $\\sigma = \\sqrt{\\frac{(b-a)^2}{12}}$\r\n\r\nFor this problem, the theoretical mean and standard deviation are\r\n\r\n$\\mu = \\frac{0+23}{2}=11.5$ seconds and $\\sigma = \\sqrt{\\frac{(23-0)^2}{12}}=6.64$ 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<\/section><\/div>\r\n

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Try It <\/span>5.2<\/span><\/h3>\r\n<\/header>
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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

\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
Table <\/span>5.2<\/span><\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/section><\/div>\r\n<\/section><\/div>\r\n
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Example <\/span>5.3<\/span><\/h3>\r\n<\/header>
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  1. Refer to Example 5.2<\/a>. What is the probability that a randomly chosen eight-week-old baby smiles between two and 18 seconds?<\/li>\r\n \t
  2. Find the 90th<\/sup> percentile for an eight-week-old baby's smiling time.<\/li>\r\n \t
  3. 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>.<\/li>\r\n<\/ol>\r\n

    Solution 5.3<\/h4>\r\n
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    1. $P(2<x<18)=\\text{(base)(height)}=(18-2)\\left(\\frac{1}{23}\\right) = \\frac{16}{23}=0.6957$<\/li>\r\n \t
    2. 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$P(x<k) = 0.90$\r\n(base)(height) = 0.90\r\n$(k-0)\\left( \\frac{1}{23}\\right) = 0.90$\r\n$k=(23)(0.90) = 20.7$\r\n\"This<\/li>\r\n \t
    3. \r\n

      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\nFind 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.\r\n\r\nWrite a new $f(x)$: $f(x) = \\frac{1}{23-8}=\\frac1{15}$ for $8<x<23$.\r\n$P(x>12 | x>8) = (23-12)\\left( \\frac1{15} \\right) = \\frac{11}{15}$\r\n\"f(X)=1\/15\r\n

      For the second way<\/strong>, use the conditional formula from Probability Topics<\/a> with the original distribution X<\/em> ~ U<\/em> (0, 23):\r\n$P(A|B) = \\frac{P(A \\text{ and } B)}{P(B)}$\r\nFor this problem, $A$ is $(x>12)$ and $B$ is $(x>8)$.<\/p>\r\nSo, $P(x>12 | x>8) = \\frac{(x>12 \\text{ and } x>8)}{P(x>8)} = \\frac{P(x>12)}{P(x>8)} = \\frac{11\/23}{15\/23} = \\frac{11}{15}$\r\n\"This<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<\/section><\/div>\r\n

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      <\/div>\r\n
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      Solution <\/span>5.3<\/span><\/h4>\r\n
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      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>) = 1<\/span><\/span>23<\/span> <\/span>\u2212<\/span> 8<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/p>\r\n123 \u2212 8123 \u2212 8