5. The police department record the speed on Campus Drive for a year and find th

Question

5. The police department record the speed on Campus Drive for a year and find that the average speed is 32.0 miles per hour with a standard deviation of 3.5. The data is approximated by a normal distribution.

A) Given a standard normal distribution (with a mean of 0 and a standard deviation of 1), find

a. P(Z < 1.51)

b. P(Z > -1.44)

c. P(-1.4 < Z < 0.6)

d. What is the value of Z if only 15.87% of all possible Z values are larger?

B) Given a normal distribution with a mean of 64 and a standard deviation of 2, find

a. P(X < 52)

b. P(X > 66)

c. P(60 < X < 68)

d. between what two X values (symmetrically distributed around the mean) are 80% of the values.

C) The long-distance calls made by the employees of a company are normally distributed with a mean of 6.3 minutes and a standard deviation of 2.2 minutes.

a. Find the probability that a call lasts between 5 and 10 minutes.

b. Find the probability that a call lasts more than 7 minutes.

c. Find the probability that a call lasts less than 4 minutes.

d. How long do the longest 10% of calls last?

Explanation / Answer

a)

P(Z<1.51) =0.9345

b)P(Z>-1.44)=0.9251

c) P(-1.4<Z<0.6)= 0.7257-0.0807 =0.6450

d) for highest 15.87% values lies at z =1

B)

a( P(X<52)=P(Z<(52-64)/2)=P(Z<-6)=0.0000

b) P(X>66)=P(Z>1 ) =0.1587

c)P(60<X<68)=P(-2<Z<2)=0.97725-0.02275 =0.9545

for 80% middle values critical z =1.28

hence 80% middle values =mean -/+z *std deviation =64 -/+1.28*2 =61.44 to 66.56

c)

a)probability that a call lasts between 5 and 10 minutes =P(5<X<10)=P(-0.5909<Z<1.6818)=0.9537-0.2773

=0.6764

b) P(X>7)=P(Z>0.3182)=0.3752

c) P(X<4) =P(Z<-1.0455)=0.1479

d) for longest 10% ; z =1.28

hence corresponding call duration =mean +z*std deviation =6.3+1.28*2.2 =9.12 minutes

Related Questions

Navigate

• Browse (All)
• Browse 5
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.