Question

A group of Brigham Young University- Idaho Students collected data of the speed of vehicles traveling through a construction zone on a state highway, where the posted construction speed limit was 25 mph. The recorded speed for 14 randomly selected vehicles is given below.

20, 24, 27, 28, 29, 30, 32, 33, 34, 36, 38, 39, 40, 40

a. determine the sample standard deviation of the speeds

b. comment on the appropriateness of using the Empirical Rule to make any general comments about the drivers speed.

c. Use the empirical rule to estimate the percentage of speeds that are between 26 and 38 mph.

d. determine the actuall percentage of drivers whose speeds are between 26 and 38 mph.

e. Use the Empirical Rule to estimate the percentage of speeds that are greater than 26 mph (so they exceed the posted speed limit)

f. determine the actual percentage of drivers whose speeds are 26 mph or higher.

Answer 1

(a)

Sample Standard deviation = ​

(b)

Empricle Diagram

Empirical rule                  
Given that the mean (u) of a normal probability distribution is 32 and the
standard deviation (sd) is 6                  
About 68% of the area under the normal curve is within one standard deviation of the mean. i.e. (u ± 1s.d)
So to the given normal distribution, about 68% of the observations lie in between
= [32 ± 6]
= [ 32 - 6 , 32 + 6]
= [ 26 , 38 ]                  
About 95% of the area under the normal curve is within two standard deviations of the mean. i.e. (u ± 2s.d)
So to the given normal distribution, about 95% of the observations lie in between
= [32 ± 2 * 6]
= [ 32 - 2 * 6 , 32 + 2* 6]
= [ 20 , 44 ]  

c.

To find P(a < = Z < = b) = F(b) - F(a)
P(X < 26) = (26-32)/6
= -6/6 = -1
= P ( Z <-1) From Standard Normal Table
= 0.02275
P(X < 38) = (38-32)/6
= 6/6 = 1
About 68% of the area under the normal curve is within one standard deviation of the mean. i.e. (u ± 1s.d)
So to the given normal distribution, about 68% of the observations lie in between
P(26 < X < 38) = 68%   

(d)

To find P(a < = Z < = b) = F(b) - F(a)
P(X < 26) = (26-32)/6
= -6/6 = -1
= P ( Z <-1) From Standard Normal Table
= 0.1587
P(X < 38) = (38-32)/6
= 6/6 = 1
= P ( Z <1) From Standard Normal Table
= 0.8413
P(26 < X < 38) = 0.8413-0.1587 = 0.6826   

(e) P(X>=26), from the empricle rule it is a boundary to the 68% of rule and above to that rule we have 13.5%+2.35 = 15.85 are percentage of drivers who were speeding.

f.

P(X >26) = 1-P(X < 26) = (26-32)/6

= -6/6 = -1
= 1-P ( Z <-1) From Standard Normal Table
= 1-0.1587

=0.8413