Understandable Statistics: Concepts and Methods
Understandable Statistics: Concepts and Methods
12th Edition
ISBN: 9781337119917
Author: Charles Henry Brase, Corrinne Pellillo Brase
Publisher: Cengage Learning
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Chapter 6.3, Problem 39P

(a)

To determine

Find the probability that the waiting time would exceed 20 minutes, given that it has exceeded 15 minutes.

(a)

Expert Solution
Check Mark

Answer to Problem 39P

The probability that the waiting time would exceed 20 minutes, given that it has exceeded 15 minutes is 0.3989.

Explanation of Solution

Calculation:

Z score:

The number of standard deviations the original measurement x is from the value of mean μ is measured using the z-score or z value. The formula for z score is,

z=xμσ

In the formula, x is the raw score, μ is the mean and σ is the standard deviation.

The conditional probability formula is,

P(A|B)=P(A and B)P(B)

The variable x denotes the length of time waiting to be seated.

The probability that the waiting time would exceed 20 minutes, given that it has exceeded 15 minutes is,

P(x>20|x>15)=P(x>20 and x>15)P(x>15)=P(x>20)P(x>15)

For P(x>20):

Substitute x as 20, μ as 18 and σ as 4 in the formula of z score.

z=20184=24=0.5

Use the Appendix II: Tables, Table 5: Areas of a Standard Normal Distribution: to obtain probability less than 0.5.

  • Locate the value 0.5 in column z.
  • Locate the value 0.00 in top row.
  • The intersecting value of row and column is 0.6915.

The probability is,

P(x>20)=P(z>0.5)=1P(z<0.5)=10.6915=0.3085

For P(x>15):

Substitute x as 15, μ as 18 and σ as 4 in the formula of z score.

z=15184=34=0.75

Use the Appendix II: Tables, Table 5: Areas of a Standard Normal Distribution: to obtain probability less than –0.75.

  • Locate the value –0.7 in column z.
  • Locate the value 0.05 in top row.
  • The intersecting value of row and column is 0.2266.

The probability is,

P(x>15)=P(z>0.75)=1P(z<0.75)=10.2266=0.7734

Substituting 0.3085 for P(x>20) and 0.7734 for P(x>15) in the required probability:

P(x>20|x>15)=0.30850.7734=0.3989

Hence, the probability that the waiting time would exceed 20 minutes, given that it has exceeded 15 minutes is 0.3989.

(b)

To determine

Find the probability that the waiting time would exceed 25 minutes, given that it has exceeded 18 minutes.

(b)

Expert Solution
Check Mark

Answer to Problem 39P

The probability that the waiting time would exceed 25 minutes, given that it has exceeded 18 minutes is 0.0802.

Explanation of Solution

Calculation:

The probability that the waiting time would exceed 25 minutes, given that it has exceeded 18 minutes is,

P(x>25|x>18)=P(x>25 and x>18)P(x>18)=P(x>25)P(x>18)

For P(x>25):

Substitute x as 25, μ as 18 and σ as 4 in the formula of z score.

z=25184=74=1.75

Use the Appendix II: Tables, Table 5: Areas of a Standard Normal Distribution: to obtain probability less than 1.75.

  • Locate the value 1.7 in column z.
  • Locate the value 0.05 in top row.
  • The intersecting value of row and column is 0.9599.

The probability is,

P(x>25)=P(z>1.75)=1P(z<1.75)=10.9599=0.0401

For P(x>18):

Substitute x as 18, μ as 18 and σ as 4 in the formula of z score.

z=18184=04=0

Use the Appendix II: Tables, Table 5: Areas of a Standard Normal Distribution: to obtain probability less than 0.

  • Locate the value 0.0 in column z.
  • Locate the value 0.00 in top row.
  • The intersecting value of row and column is 0.5000.

The probability is,

P(x>18)=P(z>0)=1P(z<0)=10.5=0.5

Substituting 0.0401 for P(x>25) and 0.5 for P(x>18) in the required probability:

P(x>25|x>18)=0.04010.5=0.0802

Hence, the probability that the waiting time would exceed 25 minutes, given that it has exceeded 18 minutes is 0.0802.

(c)

To determine

Show that P(x>20|x>15)=P(x>20)P(x>15).

(c)

Expert Solution
Check Mark

Explanation of Solution

Calculation:

Take event A as x>20 and event B as x>15 in the conditional probability formula.

P(x>20|x>15)=P(x>20 and x>15)P(x>15)

It is clear that the x is greater than 20 and x is greater than 15 this implies that the value of x is greater than 20. That is,

P(x>20|x>15)=P(x>20)P(x>15)

Hence, it is shown that P(x>20|x>15)=P(x>20)P(x>15).

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Chapter 6 Solutions

Understandable Statistics: Concepts and Methods

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