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Semenov [28]
3 years ago
13

The following data represent exam scores in a statistics class taught using traditional lecture and a class taught using a "flip

ped" classroom. Complete parts (a) through (c) below.
Traditional 70.3 68.5 79.9 67.6 85.4 79.1 56.0 80.4 79.8 71.7 63.1 70.4 59.1 75.5 71.7 63.5 71.4 77.7 92.4 79.6 77.5 83.0 69.292.6 78.0 76.9 Flipped
(a) Which course has more dispersion in exam scores using the range as the measure of dispersion'? The traditional course has a range ofwhile the "Tipped" course has a range of The Type integers or decimals. Do not round.) course has more dispersion.
(b) Which course has more dispersion in exam scores using the sample standard deviation as the measure of dispersion? The traditional course has a standard deviation of, while the "flipped" course has a standard deviation of. The (Round to three decimal places as needed.) course has more dispersion.
(c) Suppose the score of 59.1 in the traditional course was incorrectly recorded as 591. How does this affect the range?

Mathematics
1 answer:
lapo4ka [179]3 years ago
7 0

The table of this question is in the attachments.

Answer and Step-by-step explanation: <u>Standard</u> <u>deviation</u> is a measure of how spread the data is from the mean. It is calculated as:

s = √∑(x - μ)² / n - 1

where μ is the mean of the set.

<u>Range</u> is the difference between the highest and lowest value of a data set.

(a) <u>Range of Traditional course</u>:

range = 80.4 - 56

range = 24.4

<u>Range of "flipped" course</u>:

range = 92.6 - 63.5

range = 29.1

Comparing ranges, the "flipped" course has more dispersion than the traditional.

(b) <u>Standard Deviation of Traditional course</u>:

mean = 71.6

s = \sqrt{\frac{(70.3 - 71.6)^{2}+...+(59.1-71.6)^{2}}{13-1}

s = 8.95

<u>Standard Deviation of "flipped" course</u>:

mean = 77.6

s = \sqrt{\frac{(75.5 - 77.6)^{2}+...+(76.9-77.6)^{2}}{13-1}

s = 8.3

Comparing standard deviation, traditional course has more dispersion.

(c) If you change one score, range for traditional will be:

range = 591 - 56

range = 535

Changing one score increase in almost 22 times the range for this category.

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