Speeds of automobiles on a certain stretch of freeway at 11:00 PM are normally distributed with mean 65 mph. Twenty percent of the cars are traveling at speeds between 55 and 65 mph. What percentage of the cars are going faster than 75 mph?

30.15% of the cars are going faster than 75 mph.

Step-by-step explanation:

We are given that Speeds of automobiles on a certain stretch of freeway at 11:00 PM are normally distributed with mean 65 mph.

Also, Twenty percent of the cars are traveling at speeds between 55 and 65 mph.

Let X = Speeds of automobiles on a certain stretch of freeway

So, X ~ Normal()

The z-score probability distribution for normal distribution is given by;

                        Z  =    ~ N(0,1)

where,   = mean speed of automobiles = 65 mph

              = standard deviation

Now, we are given that Twenty percent of the cars are traveling at speeds between 55 and 65 mph, that means;

P(55 mph < X < 65 mph) = 0.20

P(55 mph < X < 65 mph) = P(X < 65 mph) - P(X 55 mph)

P(X < 65 mph) = P( < ) = P(Z < 0) = 0.50  {using z table}

P(X 65 mph) = P( ) = P(Z ) = 1 - P(Z < )

SO, P(55 mph < X < 65 mph) = 0.20

      P(X < 65 mph) - P(X 55 mph) = 0.20

        0.50 - [1 - P(Z < ) ] = 0.20

        P(Z < ) - 0.50 = 0.20

        P(Z < )  =  0.70

Now, looking at the z table we find that the critical value of x which gives a probability area less than 70% is 0.5244, that means;

                     =  0.5244

                    = 19.07 mph

Therefore, standard deviation, = 19.07 mph.

Now, percentage of the cars that are going faster than 75 mph is given by = P(X > 75 mph)

       P(X > 75 mph) = P( > ) = P(Z > 0.52) = 1 - P(Z 0.52)

                                                            = 1 - 0.6985 = 0.3015

Therefore, 30.15% of the cars are going faster than 75 mph.

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