what is the exact mean and standard deviation of this data 78.6
77.5
75.5
73.1
76.9



To calculate the exact mean and standard deviation of the data set:

### Data Set:
- \( 78.6, 77.5, 75.5, 73.1, 76.9 \)

### Step 1: Calculate the Mean
The mean (\( \mu \)) is given by:
\[
\mu = \frac{\sum x}{n}
\]
where \( x \) is each data point, and \( n \) is the number of data points.

\[
\mu = \frac{78.6 + 77.5 + 75.5 + 73.1 + 76.9}{5} = \frac{381.6}{5} = 76.32
\]

### Step 2: Calculate the Variance
The variance (\( \sigma^2 \)) is given by:
\[
\sigma^2 = \frac{\sum (x - \mu)^2}{n}
\]
1. Calculate \( (x - \mu)^2 \) for each data point:

   - \( (78.6 - 76.32)^2 = 5.1984 \)
   - \( (77.5 - 76.32)^2 = 1.3924 \)
   - \( (75.5 - 76.32)^2 = 0.6724 \)
   - \( (73.1 - 76.32)^2 = 10.3684 \)
   - \( (76.9 - 76.32)^2 = 0.3364 \)

2. Sum these squared deviations:
\[
5.1984 + 1.3924 + 0.6724 + 10.3684 + 0.3364 = 17.968
\]

3. Divide by \( n = 5 \):
\[
\sigma^2 = \frac{17.968}{5} = 3.5936
\]

### Step 3: Calculate the Standard Deviation
The standard deviation (\( \sigma \)) is the square root of the variance:
\[
\sigma = \sqrt{3.5936} \approx 1.896
\]

### Final Answer
- **Mean:** \( 76.32 \)
- **Standard Deviation:** \( 1.896 \)

Would you like further details on any of these calculations? 

#### Related Questions
1. How would these calculations change if we added more data points?
2. How is the mean affected by an outlier in the data set?
3. What is the difference between population and sample standard deviation?
4. How would you calculate variance if this was a sample, not a population?
5. How can standard deviation be used to understand data spread?

#### Tip
The mean is sensitive to extreme values, so it's good to check for outliers when calculating it.

What is the exact mean and standard deviation of this data 78.6, 77.5, 75.5, 73.1, 76.9

Learn how to calculate the exact mean and standard deviation of the data set 78.6, 77.5, 75.5, 73.1, 76.9. This solution walks through each step, covering core statistical concepts like mean, variance, and standard deviation, suitable for students in Grades 10-12.

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