Question

In: Statistics and Probability

Is lack of sleep causing traffic fatalities? A study found that the average number of fatal...

Is lack of sleep causing traffic fatalities? A study found that the average number of fatal crashes caused by drowsy drivers each year is 1550. Assume that annual number of fatal crashes is normally distributed with a standard deviation of 300.
a. What is the probability of fewer than 1000 fatal crashes a year due to drowsy drivers?
b. What is the probability of less than or equal to 1000 fatal crashes a year due to drowsy drivers?
c. Explain why your answers to parts a and b are equal.
d. What is the probability the number of fatal crashes will be between 1000 and 2000 for a year?
e. What is the probability of more than 1750 fatal crashes a year due to drowsy drivers?
f. What is the probability of no more than 1600 fatal crashes a year due to drowsy drivers?
g. Given that the probability of there being more than 1700 crashes per year is 0.31, use this information (not an Excel function) to find the probability of there being less than 1400 crashes per year due to drowsy drivers.
h. Explain how you determined your answer in part g using the given probability.
i. For a year to be in the upper 5% with respect to the number of fatal crashes, how many fatal crashes would have to occur?
j. Within what range evenly distributed around the mean are the number of fatal crashes that occur in 75% of the years?
lower: upper:

Solutions

Expert Solution

Solution:

   We are given that : the average number of fatal crashes caused by drowsy drivers each year is 1550 and that annual number of fatal crashes is normally distributed with a standard deviation of 300.

Thus    and

Part a. What is the probability of fewer than 1000 fatal crashes a year due to drowsy drivers?

That is : P( X < 1000 ) = ........?

Thus find z score for x = 1000

z score formula :

Thus we get : P( X < 1000 ) = P( Z < -1.83 )

Look in z table for z = - 1.8 and 0.03 and find area.

P( Z < -1.83) = 0.0336

Thus

P( X < 1000 ) = P( Z < -1.83 )

P( X < 1000 ) = 0.0336

Part b. What is the probability of less than or equal to 1000 fatal crashes a year due to drowsy drivers?

That is :

Thus find z score for x = 1000

z score formula :

Thus we get :

Look in z table for z = - 1.8 and 0.03 and find area.

From part a) we have

Thus

Part c) Explain why your answers to parts a and b are equal.

Since Normal distribution is continuous probability distribution.

Continuous probability distribution is right continuous.

That is :

That is :

Part d.What is the probability the number of fatal crashes will be between 1000 and 2000 for a year?

That is : P( 1000 < X < 2000 ) = ................?

z score for x = 1000 is z = -1.83  

and P( Z < -1.83) = 0.0336

z score for x = 2000 is :

Look in z table for z = 1.5 and 0.00 and find area.

P( Z < 1.50) = 0.9332

Thus

P( 1000 < X < 2000 ) = P( -1.83   < Z < 1.50)

P( 1000 < X < 2000 ) = P( Z < 1.50) - P( Z < -1.83)

P( 1000 < X < 2000 ) = 0.9332 - 0.0336

P( 1000 < X < 2000 ) = 0.8996


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