A teacher wants to compare how two classes performed. He records each student's final percentage in the tables below.

Class A			
40	
41	
43
43
34
84
98	
94
92	
71	
94	
85
98	
62	
86	
​
Class B			
66
39	
40	
60
61	
69	
65	
87
98	
62	
49
67
81	
93	
53	
​
a) Without computing the mean, say which class has the higher mean and why you think so.

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

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

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

Class A has the higher mean because the numbers center around a lower value than those of Class B.
​
b) Compute the mean for each class. You may round to two decimal places.

    i) Class A
Mean
Preview
​​
    ii) Class B
Mean
Preview
​
c) Without computing the standard deviation for each class, say which class has the higher standard deviation and why you think so.

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

Class B has the higher standard deviation because the spread is less than that of Class A.

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

Class A has the higher standard deviation because the spread is less than that of Class B.
​
d) Compute the standard deviation for each class. You may round to two decimal places.

    i) Class A
Standard Deviation
Preview
​
    ii) Class B
Standard Deviation
Preview

### 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.**

---

### 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.

---

### 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.**

---

### 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.

---

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.

A teacher wants to compare how two classes performed. He records each student's final percentage in the tables below.

Class A
40, 41, 43, 43, 34, 84, 98, 94, 92, 71, 94, 85, 98, 62, 86

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

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

b) Compute the mean for each class. You may round to two decimal places.

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

d) Compute the standard deviation for each class. You may round to two decimal places.

Analyze the performance of two classes using key statistical concepts like mean and standard deviation. This problem helps to compare Class A and Class B by computing their means and understanding the spread of data through standard deviation. Suitable for Grades 9-12 students.

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