# Assume the speed of vehicles along a stretch of I-10 has an approximately normal distribution with a mean of 71 mph and a standard deviation of 8 mph. a.The current speed limit is 65 mph. What is the proportion of vehicles less than or equal to the speed limit?

1. Assume the speed of vehicles along a stretch of I-10 has an approximately normal distribution with a mean of 71 mph and a standard deviation of 8 mph.

a.The current speed limit is 65 mph. What is the proportion of vehicles less than or equal

to the speed limit?

b.What proportion of the vehicles would be going less than 50 mph?

1. A new speed limit will be initiated such that approximately 10% of vehicles will be over

the speed limit. What is the new speed limit based on this criterion?

1. In what way do you think the actual distribution of speeds differs from a normal distribution?

Ans.

1. m= 71 mph; s = 8 mph

The proportion of vehicles less than or equal to the speed limit 65 mph = 0.2266

1. The proportion of the vehicles would be going less than 50 mph = 0.0043
1. The new speed limit would be 81.25 mph (approximately 81 mph).
1. The actual distribution of speeds on a highway differs from the normal distribution in that
1. It is not symmetric. The speeds are not symmetrically distributed about the central vaue. Two speeds that are an equal amount greater or lower than the mean speed, may not have the same frequency as is required for a normal distribution.
1. Secondly, the speeds further away from the average speed do not drop in frequency as rapidly as in normal distribution. The frequency of the extreme values may not be as low as in normal distribution, resulting in a fatter tail. The distribution may also be skewed in one direction. This is especially true in highways where the distribution will be expected to be positively skewed- greater frequency in the higher speed range than lower.
1. A group of students at a school takes a history test. The distribution is normal with a mean of 25, and a standard deviation of 4. (a) Everyone who scores in the top 30% of the distribution gets a certificate. What is the lowest score someone can get and still earn a certificate? (b) The top 5% of the scores get to compete in a statewide history contest. What is the lowest score someone can get and still go onto compete with the rest of the state?

Ans. For the test, the mean score m= 25; s = 4

Using the normal calculator (Ch-7, Lane),

1. the lowest score someone can get and still earn a certificate = 27.096 (or approx.. 27.1)
1. the lowest score someone can get and still go onto compete with the rest of the state = 31.58 (or approx. 31.6)
1. Use the normal distribution to approximate the binomial distribution and find the probability of getting 15 to 18 heads out of 25 flips. Compare this to what you get when you calculate the probability using the binomial distribution. Write your answers out to four decimal places.

Ans  For the binomial distribution of coin flipping, probability of getting a head (success) in one flip = p = 0.5

N = no. of trials = 25

Mean is m = N*p = 25 * 0.5 = 12.5

Variance ?2 = Np(1-p) = (25)(0.5)(0.5) = 6.25

The standard deviation is therefore ? = ?6.25 = 2.5.

Using the binomial calculator (Lane CH-5),

the probability of getting 15 to 18 heads out of 25 flips = 0.2049

Using the normal distribution to approximate the binomial distribution: The area between 14.5 and 18.5 in the normal distribution curve is an approximation of the probability of obtaining 15 to 18 heads.

Area below 14.5 = 0.7881

Area below 18.5 = 0.9918

Therefore,

Area between 14.5 and 18.5 = 0.9918 – 0.7881 = 0.2037

So in this case,

the probability of getting 15 to 18 heads out of 25 flips = 0.2037

Illowsky CH-6

1. The patient recovery time from a particular surgical procedure is normally distributed with a mean of 5.3 days and a standard deviation of 2.1 days.

What is the median recovery time?

1. 2.7
1. 5.3
1. 7.4
1. 2.1

Ans. b. 5.3

In the normal distribution, mean and median are the same.

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