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.