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GREYUIT [131]
2 years ago
12

The table shows the mean monthly temperatures in Alcudia for the last 2 years in degrees Celsius. °C Jan Feb Mar Apr May Jun Jul

Aug Sep Oct Nov Dec Last Year 9 10 11 14 19 22 27 29 21 18 11 6 2 Years Ago 7 11 10 13 17 23 25 25 20 18 12 9. What was the modal temperature last year?
Mathematics
1 answer:
gtnhenbr [62]2 years ago
8 0

Answer:

<u>The modal temperature last year is 11 (March and November)</u>

Step-by-step explanation:

1. Let's review the information given to us to answer the question correctly:

A year ago Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec

                     9    10     11   14    19     22  27  29   21    18    11   6

2 years ago Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec

                      7     11     10   13    17     23  25  25    20   18  12  9

2. What was the modal temperature last year?

Let's recall the mode of a data set is the number that occurs most frequently in the set. The number that occurs the most in the average temperatures last year is 11 (March and November) and 2 years ago is 25 (July and August)                  

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Answer:

a) The percentage of athletes whose GPA more than 1.665 is 87.49%.

b) John's GPA is 3.645.

Step-by-step explanation:

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

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a)Find the percentage of athletes whose GPA more than 1.665.

This is 1 subtracted by the pvalue of Z when X = 1.665. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{1.665 - 2.7}{0.9}

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1 - 0.1251 = 0.8749

The percentage of athletes whose GPA more than 1.665 is 87.49%.

b) John's GPA is more than 85.31 percent of the athletes in the study. Compute his GPA.

His GPA is X when Z has a pvalue of 0.8531. So it is X when Z = 1.05.

Z = \frac{X - \mu}{\sigma}

1.05 = \frac{X - 2.7}{0.9}

X - 2.7 = 1.05*0.9

X = 3.645

John's GPA is 3.645.

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