For the past 20 years, the high temperature on April 15th has averaged µ = 62 degrees with a standard deviation of σ = 4. Last year, the high temperature was 72 degrees. Based on this information, last year's temperature on April 15th was the following.

a. far above average
b. above average, but it is impossible to describe how much above average
c. There is not enough information to compare last year with the average.
d. a little above average

2.5 standard deviations above average is far above average. That number implies a probability of 0.4% that high temperatre is 72 or above.

far above average

a.far above average

ERE

For the past 20 years, the high temperature on April 15th has averaged ì = 62 degrees with a standard deviation of ó = 4. Last year, the high temperature was 72 degrees. Based on this information, last year's temperature on April 15th was the following

Well, last year's temperature on April 15th was as hot as a freshly roasted marshmallow! So, I'm going to go with option a. far above average. I hope you brought some sunscreen!

To determine how last year's temperature on April 15th compares to the average, we can use the concept of z-scores.

A z-score measures how many standard deviations an observation is from the mean. We can calculate the z-score of last year's temperature using the formula:

z = (x - µ) / σ

Where:
- x is the value of the observation (last year's temperature)
- µ is the mean of the distribution (average temperature)
- σ is the standard deviation

In this case, the mean (µ) is 62 and the standard deviation (σ) is 4. Last year's temperature was 72.

Substituting the values into the formula, we get:

z = (72 - 62) / 4 = 10 / 4 = 2.5

The z-score of 2.5 tells us that last year's temperature is 2.5 standard deviations above the average.

Now, to answer the question:

a. far above average: Since a z-score of 2.5 indicates that the observation is significantly above the mean, we can say that last year's temperature was far above average.

Therefore, the correct answer is a. far above average.