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3 years ago

Francis analyzes the given data sets.

Mathematics

1 answer:

Anit [1.1K]3 years ago

5 0

Answer:

2. The IQR for Data Set A is larger than the IQR for Data Set B.

4. The IQR for Data Set A and Data Set B measures the middle 50% of the data.

Step-by-step explanation:

IQR is an abbreviated form of the word Interquartile range. It is used as a measure in statistics for a set of given data.

Interquartile range can be defined as the 50th percentile (50%) of a set of data. It is the difference between the upper limit and lower limit of a given set of data.

Interquartile range can also be referred to or known as the Median(middle) of a given set of data.

Interquartile range is determined by arrange the given set of data from the lowest to height number and then we find or pick the number that is in the middle as our Interquartile range. If the number of term of a given set of data is even in number, we pick two numbers in the middle, add them together and divide it by 2.

In the question above, the correct statements that Francis can make about a given set of data is :

2. "The IQR for Data Set A is larger than the IQR for Data Set B" and

4. The IQR for Data Set A and Data Set B measures the middle 50% of the data.

This is because

Data Set A: 20, 25, 28, 30, 39, 40

Since the number of terms in data set A is 6 (i.e an even number) hence, we pick the two numbers in the middle

The Interquartile range for Data Set A= (28+30) ÷ 2 = 58÷ 2 = 29

For Data Set B: 20, 22, 23, 27, 30, 42

The number of terms in data set is 6( i.e an even number), likewise we pick the two numbers in the middle

The Interquartile range for Data Set B

= (23+27) ÷ 2 = 50÷ 2 = 25

From the above calculation, we have proved that option 2 is correct. The Interquartile range of Data Set A which is 29 is larger than the Interquartile range of Data Set B which is 25.

Likewise Option 4 is correct as well , because the Interquartile range of both Data set A and Data set B measured the middle 50% of the data.

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$\frac{(6+2)^3-12}{5}\\\\\frac{(8)^3-12}{5}\\\\\frac{512-12}{5}\\\\\frac{500}{5}\\\\100\\\\$

So in short, $\frac{(6+2)^3-12}{5}=100\\\\$

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The mistake Jerry likely made was that he only cubed the 2 and didn't realize the 6 was part of that cubing process. It seems he didn't add first and decided to cube before adding.

This is probably what steps Jerry did

$\frac{6+2^3-12}{5}\\\\\frac{6+8-12}{5}\\\\\frac{14-12}{5}\\\\\frac{2}{5}\\\\$

But as mentioned, those steps are incorrect because the 6 is part of the cubing operation. In other words, Jerry should have added the 6+2 first before cubing afterward (due to PEMDAS determining the order of operations).

Or you could think of it like this

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