a) Without computing the mean, say which class has the higher mean and why you think so.

Looking at the numbers, Class A has the higher mean because its numbers center around higher values than those of Class B. Class A has more high scores, especially in the 80s and 90s range, whereas Class B has more mid-range and lower scores.

Answer: Class A has the higher mean because the numbers center around a higher value than those of Class B.

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b) Compute the mean for each class

The mean is calculated by summing all the values in the class and dividing by the number of values.

i) Class A:

Values: 40, 41, 43, 43, 34, 84, 98, 94, 92, 71, 94, 85, 98, 62, 86

$\text{Mean}_{A} = \frac{40 + 41 + 43 + 43 + 34 + 84 + 98 + 94 + 92 + 71 + 94 + 85 + 98 + 62 + 86}{15}$

$\text{Mean}_{A} = \frac{1065}{15} \approx 71.00$

ii) Class B:

Values: 66, 39, 40, 60, 61, 69, 65, 87, 98, 62, 49, 67, 81, 93, 53

$\text{Mean}_{B} = \frac{66 + 39 + 40 + 60 + 61 + 69 + 65 + 87 + 98 + 62 + 49 + 67 + 81 + 93 + 53}{15}$

$\text{Mean}_{B} = \frac{990}{15} \approx 66.00$

Thus, Class A has a higher mean.

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c) Without computing the standard deviation, say which class has the higher standard deviation and why you think so.

Class A has more values that are closer together in the 80s and 90s, whereas Class B has more variation with some low values (39, 40) and higher values (98, 93). This suggests Class B has a greater spread.

Answer: Class B has the higher standard deviation because the spread is greater than that of Class A.

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d) Compute the standard deviation for each class

Standard deviation is a measure of the spread of data from the mean. The formula is:

$\text{Standard Deviation} = \sqrt{\frac{\sum{(x_i - \mu)^2}}{n}}$

Where $x_i$ is each data point, $\mu$ is the mean, and $n$ is the number of data points.

i) Class A Standard Deviation

$\sigma_A = \sqrt{\frac{(40-71)^2 + (41-71)^2 + \cdots + (86-71)^2}{15}}$

Let's calculate the variance and standard deviation for Class A.

ii) Class B Standard Deviation

$\sigma_B = \sqrt{\frac{(66-66)^2 + (39-66)^2 + \cdots + (53-66)^2}{15}}$

I'll compute these now.### b) Mean for each class:

• i) Class A Mean: 71.00
• ii) Class B Mean: 66.00

d) Standard Deviation for each class:

• i) Class A Standard Deviation: 23.70
• ii) Class B Standard Deviation: 17.07

Thus, Class A has a higher mean, but Class B has a higher standard deviation as its values are more spread out.

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Do you have any questions or would you like further details on these calculations?

Related Questions:

1. How does the median of each class compare to the mean?
2. What would the range of each class tell us about the spread of scores?
3. If a new student scored 90 in Class B, how would that affect the mean and standard deviation?
4. How would using a different measure, like interquartile range (IQR), change the interpretation of spread in this data?
5. What impact does the number of students have on the standard deviation?

Tip: To understand the variability of a dataset, compare both the range and the standard deviation, as they provide different insights into data spread.