MATHEMATICS HIGH SCHOOL

The average high temperature in Anchorage, Alaska, in January is 21°F with a standard deviation of 10°. The average high temperature in Honolulu in January is 80°F with a standard deviation of 8°. In which location would it be more unusual to have a day in January with a high of 57°F?

Answers

Answer 1

Answer:

Answer:

In Honolulu it be more unusual to have a day in January with a high of 57°F

Step-by-step explanation:

Given,

The average high temperature in Anchorage, Alaska,

$\mu_1$ = 21° F,

Having standard deviation,

$\sigma_1$ = 10°,

Thus, the z-score for the high temperature of 57° F in alaska,

$z_1=(57-\mu_1)/(\sigma_1)$

$=(57-21)/(10)$

$=3.6$

By the z-score table the probability of the high temperature of 57°F in Alaska = 0.99984 = 99.984 %

Now, the average high temperature in Honolulu is,

$\mu_2=80^(\circ)$

Having standard deviation,

$\sigma_2=8^(\circ)$

Thus, the z-score for the high temperature of 57° F in Honolulu,

$z_2=(57-\mu_2)/(\sigma_2)$

$=(57-80)/(8)$

$=-2.875$

By the z-score table the probability of the high temperature of 57° in Honolulu = 0.00205 = 0.205 %

Since, a probability is unusual if it is less than 5 %,

Hence, In Honolulu it be more unusual to have a day in January with a high of 57°F.

Answer 2

Answer: It would be more odd in Honolulu.

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Answers

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