Question

The distribution of passenger vehicle speeds traveling on a certain freeway in California is nearly normal with a mean of 71.5 miles/hour and a standard deviation of 4.75 miles/hour. The speed limit on this stretch of the freeway is 70 miles/hour. (a) A highway patrol officer is hidden on the side of the freeway. What is the probability that 5 cars pass and none are speeding? Assume that the speeds of the cars are independent of each other. (Round your answer to four decimal places.) .0074 (b) On average, how many cars would the highway patrol officer expect to watch until the first car that is speeding? (Round your answer to two decimal places.) What is the standard deviation of the number of cars he would expect to watch? (Round your answer to two decimal places.)

Answer 1

Answer:

a

$G = 0.007523$

b

 The  number of cars the  highway patrol officer would watch before a  car that is seen is $E(X) = 1.6027$

The standard deviation is   $s = 0.9829$

gg

Step-by-step explanation:

From the question we are told that

The mean is $\mu = 71.5 \ miles/hour$

The standard deviation is $\sigma = 4.75 \ miles/hour$

The speed limit is $x = 70 \ miles /hour$

Generally the probability of getting a car that is moving with speed greater than the speed limit is mathematically represented as

$p =P(X > x ) = P(X > 70) = P(\frac{X - \mu }{\sigma } > \frac{70 - 71.5 }{4.75})$

=> $p= P(X > 70) = P(\frac{X - \mu }{\sigma } > \frac{70 - 71.5 }{4.75})$

=> $p= P(X > 70) = P(\frac{X - \mu }{\sigma } > -0.31579 )$

Here

$\frac{X - \mu }{\sigma } = Z(The \ standardized \ value \ of X )$

So

=> $p= P(X > 70) = P(Z > -0.31579 )$

From the z-table

$p = P(Z > -0.31579 ) = 0.62392$

So

$p = P(X > 70) = 0.62392$

Generally the probability of getting a car that is not moving with speed greater than the speed limit is mathematically represented as

$q = 1 - p$

=> $q = 1 - 0.62392$

=> $q = 0.37608$

Generally the probability of getting 5 cars that are not speeding is mathematically represented as

$G = q^5$

=> $G = (0.37608)^5$

=> $G = 0.007523$

Generally the number of cars that the highway patrol officer is expected to watch until the first car that is speeding is gotten is mathematically represented as

$E(X) = \frac{1}{p}$

=> $E(X) = \frac{1}{0.62392}$

=> $E(X) = 1.6027$

Generally the standard deviation is mathematically represented as

$s = \sqrt{\frac{1 - p }{ p^2} }$

=>     $s = \sqrt{\frac{1 -0.62392 }{ (0.62392)^2} }$

=> $s = 0.9829$