Question
Christine is currently taking a college astronomy class and the instructor often gives quizzes. On the past seven quizzes, Christine got the scores shown below. Find the standard deviation, rounding to one more decimal place than is present in the original data
Answers
GPT-4o mini
To find the standard deviation of Christine's quiz scores, you first need to calculate the mean (average) of the scores, then find the squared differences from the mean, calculate the variance, and finally, find the standard deviation.
Let's assume Christine's quiz scores are:
\( x_1, x_2, x_3, x_4, x_5, x_6, x_7 \)
We will denote these scores. For the sake of this example, let's say the scores are: 82, 90, 76, 88, 92, 85, and 78. Please replace these with the actual scores if they are different.
### Step 1: Calculate the Mean
\[
\text{Mean} (\mu) = \frac{x_1 + x_2 + x_3 + x_4 + x_5 + x_6 + x_7}{7}
\]
\[
\mu = \frac{82 + 90 + 76 + 88 + 92 + 85 + 78}{7} = \frac{511}{7} \approx 73
\]
### Step 2: Calculate the Squared Differences
Next, calculate the squared differences from the mean for each score.
\[
(x_i - \mu)^2
\]
1. \( (82 - 85)^2 = 9 \)
2. \( (90 - 85)^2 = 25 \)
3. \( (76 - 85)^2 = 81 \)
4. \( (88 - 85)^2 = 9 \)
5. \( (92 - 85)^2 = 49 \)
6. \( (85 - 85)^2 = 0 \)
7. \( (78 - 85)^2 = 49 \)
### Step 3: Calculate the Variance
Now, calculate the average of those squared differences. This average is the variance. Since we are dealing with a sample, we use \( n-1 \) for the denominator.
\[
\text{Variance} = \frac{\sum (x_i - \mu)^2}{n-1} = \frac{9 + 25 + 81 + 9 + 49 + 0 + 49}{6} = \frac{222}{6} = 37
\]
### Step 4: Calculate the Standard Deviation
The standard deviation is the square root of the variance.
\[
\sigma = \sqrt{\text{Variance}} = \sqrt{37} \approx 6.0827
\]
### Step 5: Round the Answer
Rounding to one more decimal place than is present in the original data (which I will assume is to one decimal place), we round \( 6.0827 \) to \( 6.1 \).
Therefore, the standard deviation of Christine's quiz scores is approximately **6.1**.
Please provide the actual scores if you want a precise calculation based on her scores.
Let's assume Christine's quiz scores are:
\( x_1, x_2, x_3, x_4, x_5, x_6, x_7 \)
We will denote these scores. For the sake of this example, let's say the scores are: 82, 90, 76, 88, 92, 85, and 78. Please replace these with the actual scores if they are different.
### Step 1: Calculate the Mean
\[
\text{Mean} (\mu) = \frac{x_1 + x_2 + x_3 + x_4 + x_5 + x_6 + x_7}{7}
\]
\[
\mu = \frac{82 + 90 + 76 + 88 + 92 + 85 + 78}{7} = \frac{511}{7} \approx 73
\]
### Step 2: Calculate the Squared Differences
Next, calculate the squared differences from the mean for each score.
\[
(x_i - \mu)^2
\]
1. \( (82 - 85)^2 = 9 \)
2. \( (90 - 85)^2 = 25 \)
3. \( (76 - 85)^2 = 81 \)
4. \( (88 - 85)^2 = 9 \)
5. \( (92 - 85)^2 = 49 \)
6. \( (85 - 85)^2 = 0 \)
7. \( (78 - 85)^2 = 49 \)
### Step 3: Calculate the Variance
Now, calculate the average of those squared differences. This average is the variance. Since we are dealing with a sample, we use \( n-1 \) for the denominator.
\[
\text{Variance} = \frac{\sum (x_i - \mu)^2}{n-1} = \frac{9 + 25 + 81 + 9 + 49 + 0 + 49}{6} = \frac{222}{6} = 37
\]
### Step 4: Calculate the Standard Deviation
The standard deviation is the square root of the variance.
\[
\sigma = \sqrt{\text{Variance}} = \sqrt{37} \approx 6.0827
\]
### Step 5: Round the Answer
Rounding to one more decimal place than is present in the original data (which I will assume is to one decimal place), we round \( 6.0827 \) to \( 6.1 \).
Therefore, the standard deviation of Christine's quiz scores is approximately **6.1**.
Please provide the actual scores if you want a precise calculation based on her scores.