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Chapter 3: Q10E (page 100)

A civil engineer is studying a left-turn lane that is long enough to hold seven cars. LetXbe the number of cars in the lane at the end of a randomly chosen red light. The engineer believes that the probability thatX=xis proportional to(x+1)(8鈭�x)forx=0, . . . ,7 (the possible values ofX).

a. Find the p.f. ofX.

b. Find the probability thatXwill be at least 5.

Short Answer

Expert verified

a. The probability function is\(P\left( {X = x} \right) = \left\{ \begin{array}{l}\frac{1}{{120}}\left( {x + 1} \right)\left( {8 - x} \right)\;\;x = 0,1,...,7\\0\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;elsewhere\end{array} \right.\)

b. Theprobability that Xwill be at least 5 is 0.33.

Step by step solution

01

Given information

The random variable X represents the number of cars in the lane at the end of a randomly selected chosen red light.

The probability of an event x of X is proportional to \(\left( {x + 1} \right)\left( {8 - x} \right)\) , where \(x = 0, \ldots ,7\) .

02

Determine the probability function of X

a.

From the provided information,

\(\begin{array}{c}P\left( {X = x} \right) \propto \left( {x + 1} \right)\left( {8 - x} \right)\\ = k\left( {x + 1} \right)\left( {8 - x} \right)\;\;\;for\;x = 0,...,7\end{array}\)

The value of k is computed using the property that the sum of probabilities over the support of a random variable is 1.

\(\begin{array}{c}\sum\limits_{x = 0}^7 {k\left( {x + 1} \right)\left( {8 - x} \right)} = 1\\k\left( {8 + 14 + 18 + 20 + 20 + 18 + 14 + 8} \right) = 1\\120k = 1\\k = \frac{1}{{120}}\end{array}\)

The probability function of X is,

\(P\left( {X = x} \right) = \left\{ \begin{array}{l}\frac{1}{{120}}\left( {x + 1} \right)\left( {8 - x} \right)\;\;x = 0,1,...,7\\0\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;elsewhere\end{array} \right.\)

03

Compute the probability

b.

The probability that X will be at least 5 is computed as,

\(\begin{array}{c}P\left( {X \ge 5} \right) = P\left( {X = 5} \right) + P\left( {X = 6} \right) = P\left( {X = 7} \right)\\ = \frac{1}{{120}}\left( {5 + 1} \right)\left( {8 - 5} \right) + \frac{1}{{120}}\left( {6 + 1} \right)\left( {8 - 6} \right) + \frac{1}{{120}}\left( {7 + 1} \right)\left( {8 - 7} \right)\\ = \frac{1}{{120}}\left( {18 + 14 + 8} \right)\\ = 0.33\end{array}\)

Therefore, the probability that X will be at least 5 is 0.33.

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