Question

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.

Answer 1

Answer:

Step-by-step explanation:

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

Given that:

For morning flights

Mean $\mu_1$ = 15

standard deviation $\sigma_1$ = 5

Sample size $n_1$ = 10

$X_1 \sim Normal ( \mu_1, \sigma _1)$

The Z - score is calculated as:

$Z = \dfrac{x_1 - \mu_1}{\sigma_1}$

$Z = \dfrac{30 -15}{5}$

$Z = \dfrac{15}{5}$

Z = 3

For evening flights

Mean $\mu_2$ = 20

standard deviation $\sigma_2$ = 3

Sample size $n_2$ = 10

$X_2 \sim Normal ( \mu_2, \sigma _2)$

$Z = \dfrac{x_2 - \mu_2}{\sigma_2}$

$Z = \dfrac{30 -20}{3}$

$Z = \dfrac{10}{3}$

Z = 3.33

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