8 Assume the speed of vehicles along a stretch of I 10 has a

8. 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 50mph? c. 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? d. In what way do you think the actual distribution of speeds differs from a normal distribution? 11. 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?

Solution

(8)(a)P(X<65) = P((X-mean)/s <(65-71)/8)

=P(Z<-0.75) = 0.2266 (from standard normal table)

i.e. 22.66%

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(b)P(X<50) = P(Z<(50-71)/8) = P(Z<-2.63) =0.0043 (from standard normal table)

i.e. 0.43%

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(c)P(X>c)=0.1

--> P(Z<(c-71)/8)=1-0.1=0.9

--> (c-71)/8 = 1.28 (from standard normal table)

So c= 71+1.28*8=81.24

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(d)If the distribution of speeds would be skewed to the right because most of vehicles would be less than the speed limit

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(a)P(X>c)=0.3

--> P(Z<(c-25)/4) = 1-0.3=0.7

--> (c-25)/4 = 0.52 (from standard normal table)

So c= 25+0.52*4=27.08

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(b)P(X>c)=0.05

--> P(Z<(c-25)/4) =1-0.05=0.95

--> (c-25)/4=1.64 (from standard normal table)

So c=25+1.64*4=31.56