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 6.5, Problem 22P

(a)

To determine

Find the probability that total taxi and take off time would be less than 320 minutes for 36 jets on a given runway.

(a)

Expert Solution
Check Mark

Answer to Problem 22P

The probability that total taxi and take off time would be less than 320 minutes for 36 jets on a given runway is 0.8238.

Explanation of Solution

Calculation:

Central limit theorem:

The characteristics of the sampling distribution x¯, for all the sample size n is determined using the central limit theorem. Based on central limit theorem the sample size n must be greater than or equal to 30 for making the distribution of sample means x¯ to be approximately normal, even if the population is not normally distributed.

Sampling distribution:

For x distribution the sampling distribution of x¯ is,

μx¯=μσx¯=σn

In the formula, μ denotes population mean, σ denotes population standard deviation, n denotes the sample size.

The x¯ can be converted into standard normal distribution as,

z=x¯μx¯σx¯

In the formula, x¯ is the mean raw score, μx¯ is the mean of the sampling distribution and σx¯ is the standard deviation of the sampling distribution.

The random variable x is defined as taxi and take-off time for commercial jets. The xi is denotes the take-off time for ith commercial jets (i=1,2,...,36). The sum of the take-off time for all the 36 commercial jets is, w=x1+x2+...+x36. The total taxi and take-off time for 36 commercial jets on runway (w) is less than 320 minutes has to be calculated which is denoted as P(w<320).

Consider the expression w<320. Divide both sides of the expression with 36.

w<320w36<32036w36<8.889

The sum of taxi and take off time for 36 commercial jets on runway is denoted by w. That is,

w36=x1+x2+...+x3636=x¯

The total taxi and take off time for 36 commercial jets on runway (w) is less than 320 can be written as,

P(w<320)=P(w36<8.889)=P(x¯<8.889)

Hence, the quantities P(w<320) and P(x¯<8.889) are equal.

The number of commercial jets on runway in the study is n=36. The sample size is greater than 36; based on the central limit theorem the distribution of sample means x¯ is approximately normal.

The mean of the x distributions is μ=8.5minutes. The mean of sampling distribution is equal to the mean of population distribution. That is,

μx¯=μ=8.5

The standard deviation of the x distributions is σ=2.5minutes. The standard deviation of x¯ distribution is,

σx¯=σn=2.536=2.56=0.4167

Z score:

Substitute x¯ as 8.889, μx¯ as 8.5 and σx¯ as 0.4167 in the formula of z score.

z=8.8898.50.4167=0.3890.4167=0.93

Use the Appendix II: Tables, Table 5: Areas of a Standard Normal Distribution: to obtain probability less than 0.93.

  • Locate the value 0.9 in column z.
  • Locate the value 0.03 in top row.
  • The intersecting value of row and column is 0.8238.

The probability is,

P(x¯<8.889)=P(z<0.93)=0.8238

Hence, the probability that total taxi and take off time would be less than 320 minutes for 36 jets on a given runway is 0.8238.

(b)

To determine

Find the probability that total taxi and take off time would be more than 275 minutes for 36 jets on a given runway.

(b)

Expert Solution
Check Mark

Answer to Problem 22P

The probability that total taxi and take off time would be more than 275 minutes for 36 jets on a given runway is 0.9808.

Explanation of Solution

Calculation:

The total taxi and take-off time for 36 commercial jets on runway (w) is more than 275 minutes has to be calculated which is denoted as P(w>275).

Consider the expression w>275. Divide both sides of the expression with 36.

w>275w36>27536w36>7.639

The sum of taxi and take off time for 36 commercial jets on runway is denoted by w. That is,

w36=x1+x2+...+x3636=x¯

The total taxi and take off time for 36 commercial jets on runway (w) is more than 275 can be written as,

P(w>275)=P(w36>7.639)=P(x¯>7.639)

Hence, the quantities P(w>275) and P(x¯>7.639) are equal.

Z score:

Substitute x¯ as 7.639, μx¯ as 8.5 and σx¯ as 0.4167 in the formula of z score.

z=7.6398.50.4167=0.8610.4167=2.07

Use the Appendix II: Tables, Table 5: Areas of a Standard Normal Distribution: to obtain probability less than –2.07.

  • Locate the value –2.0 in column z.
  • Locate the value 0.07 in top row.
  • The intersecting value of row and column is 0.0192.

The probability is,

P(x¯>7.639)=P(z>2.07)=1P(z2.07)=10.0192=0.9808

Hence, the probability that total taxi and take off time would be more than 275 minutes for 36 jets on a given runway is 0.9808.

(c)

To determine

Find the probability that total taxi and take off time would be between 275 minutes and 320 minutes for 36 jets on a given runway.

(c)

Expert Solution
Check Mark

Answer to Problem 22P

The probability that total taxi and take off time would be between 275 minutes and 320 minutes for 36 jets on a given runway is 0.8046.

Explanation of Solution

Calculation:

From part (a), the quantity 320 minutes is equal to 8.889 after conversion. The z score is 0.93. The probability value for less than 0.93 from normal table is 0.8238.

From part (b), the quantity 275 minutes is equal to 7.639 after conversion. The z score is 2.07. The probability value for less than –2.07 from normal table is 0.0192.

The probability is,

P(275<w<320)=P(7.639<x¯<8.889)=P(2.07<z<0.93)=P(z<0.93)P(z<2.07)=0.82380.0192=0.8046

Hence, the probability that total taxi and take off time would be between 275 minutes and 320 minutes for 36 jets on a given runway is 0.8046.

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Understandable Statistics: Concepts and Methods

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