Question

The previous exercises states that the distribution of speeds of cars traveling on the Interstate 5 Freeway (I-5) in California is nearly normal with a mean of 72.6 miles/hour and a standard deviation of 4.78 miles/hour. The speed limit on this stretch of the I-5 is 70 miles/hour. 5. 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.

Answer 1

Answer:

The probability is $P(X \le 70 ) = 0.7054$

Step-by-step explanation:

From the question we are told that

The mean is $\mu = 72.6 \ miles /hour$

The standard deviation is $\sigma = 4.78 \ miles /hour$

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

Generally the probability that 5 cars pass and none are speeding is equivalent to the probability that the speed of the cars is less than x which is mathematically represented as

$P(X \le x ) = 1 - P(X > x )$

Now

$P(X > x) = P(\frac{X - \mu}{\sigma } >\frac{x - \mu}{\sigma } )$

Generally  

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

So

    $P(X > 70) = P(Z >\frac{70 - 72.6}{72.6 } )$

      $P(X > 70) = P(Z > -0.54 )$

From the z-table  

           $P(Z > -0.54 ) = 0.2946$

=>      $P(X > 70) = 0.2946$

So

    $P(X \le 70 ) = 1 - P(X > 70 )$

=>  $P(X \le 70 ) = 1 - 0.2946$

=>  $P(X \le 70 ) = 0.7054$