An airline has two flights each day from City A to City B. A random sample of 10 morning flights left the gate an average of 15 minutes late with a standard deviation of 5 minutes. A random sample of 10 evening flights left the gate an average of 20 minutes late with a standard deviation of 3 minutes. Suppose both the morning and evening flights are 30 minutes late. To determine which flight is more late than usual, first find the z-scores. Round to one decimal place if necessary.

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Answer:

Step-by-step explanation:

Let X denote the random variable that obeys the normal distribution.

Given that:

For morning flights

Mean [tex]\mu_1[/tex] = 15

standard deviation [tex]\sigma_1[/tex] = 5

Sample size [tex]n_1[/tex] = 10

[tex]X_1 \sim Normal ( \mu_1, \sigma _1)[/tex]

The Z - score is calculated as:

[tex]Z = \dfrac{x_1 - \mu_1}{\sigma_1}[/tex]

[tex]Z = \dfrac{30 -15}{5}[/tex]

[tex]Z = \dfrac{15}{5}[/tex]

Z = 3

For evening flights

Mean [tex]\mu_2[/tex] = 20

standard deviation [tex]\sigma_2[/tex] = 3

Sample size [tex]n_2[/tex] = 10

[tex]X_2 \sim Normal ( \mu_2, \sigma _2)[/tex]

[tex]Z = \dfrac{x_2 - \mu_2}{\sigma_2}[/tex]

[tex]Z = \dfrac{30 -20}{3}[/tex]

[tex]Z = \dfrac{10}{3}[/tex]

Z = 3.33

Hence, from the above z-scores, we will realize that the evening flight is more late than usual.

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