3. Information from the Department of Motor Vehicles indicates that the average

Question

3. Information from the Department of Motor Vehicles indicates that the average age of licensed drivers is -457 years with a standard deviation of ?-125 years. Assuming that the distribution of divers' ages is approx following questions: imately normal, answer the a. What proportion of licensed drivers are older than 50 years old? (12.5 points) b. What proportion of licensed drivers are younger than 30 years old? (12.5 points) 4. A report in 2010 indicates that Americans between the ages of 8 and 18 spend an average of 7.5 ours per day using some sort of electronic device such as smart phones, computers, or tablets. Assume that the distribution of times is normal with a standard deviation of o 2.5 hours and find the following values: a. What is the probability of selecting an individual who uses electronic devices more than 12 hours a day? (12.5 points) b. What proportion of 8 to 18 year-old Americans spend between 5 and 10 hours per day using electronic devices. In symbols it would look like this except the less than symbol SHOULD NOT HAVE THE UNDERLINE MAKING IT LESS THAN OR EQUAL TO (couldn't edit the document to get rid of the line underneath, so the symbol version should just say p[S less than X less than 1) (ie., .p/5 s XS 10 2) (12.5 points) B Send to Binder Download Print Open with docReader Activity Details You have viewed this topic O Due March 25 at 11.30 EM Ends Mar 28, 2018 14.30PM Last Visited Mar 25, 2018 9:47 PM >a

Explanation / Answer

3)
mean = 45.7 , s = 12.5

a)
P(x >50)
By normal distribution,

z = (x -mean)/s
= ( 50 - 45.7)/12.5
= 0.344
P(x >50) = P(z > 0.344) = 0.3654 by using standard normal table

b)
P(x <30)
By normal distribution,

z = (x -mean)/s
= ( 30 - 45.7)/12.5
= -1.256
P(x < 30) = P(z < -1.256) = 0.1046 by using standard normal table

4)
mean = 7.5 , s = 2.5

a)
P(x >12)
By normal distribution,

z = (x -mean)/s
= ( 12 - 7.5)/2.5
= 1.8
P(x >12) = P(z > 1.8) = 0.0359 by using standard normal table

b)
P(5 < x < 10)
= P((5 - 7.5)/2.5 < z < (10 - 7.5) /2.5)
= P(-1 < z <1)
P(5 < x < 10) = P(-1 < z <1) = 0.6827by using standard normal table

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