On 6 quizzes Joe scored 73,88,86,90, 87 and 29. What is the mean and standard deviation. On the 7th quiz he scored 32 - how does this affect the mean and standard deviation?

Let's calculate the mean and standard deviation for Joe's initial 6 quiz scores, and then see how they change after his 7th quiz score.

### 1. Initial Mean and Standard Deviation (6 quizzes)
Joe's scores: 73, 88, 86, 90, 87, 29

**Mean (Average):**
\[
\text{Mean} = \frac{73 + 88 + 86 + 90 + 87 + 29}{6} = \frac{453}{6} = 75.5
\]

**Standard Deviation:**
The standard deviation is calculated using the formula:
\[
\text{Standard Deviation} = \sqrt{\frac{\sum (x_i - \mu)^2}{N}}
\]
where:
- \(x_i\) is each individual score,
- \(\mu\) is the mean,
- \(N\) is the number of scores.

Let's compute the variance first:
\[
\begin{align*}
(73 - 75.5)^2 &= 6.25 \\
(88 - 75.5)^2 &= 156.25 \\
(86 - 75.5)^2 &= 110.25 \\
(90 - 75.5)^2 &= 210.25 \\
(87 - 75.5)^2 &= 132.25 \\
(29 - 75.5)^2 &= 2162.25 \\
\end{align*}
\]

Sum of squared deviations:
\[
\text{Sum} = 6.25 + 156.25 + 110.25 + 210.25 + 132.25 + 2162.25 = 2777.5
\]

Variance:
\[
\text{Variance} = \frac{2777.5}{6} = 462.92
\]

Standard deviation:
\[
\text{Standard Deviation} = \sqrt{462.92} \approx 21.52
\]

### 2. After the 7th Quiz Score (32)

Joe's new scores: 73, 88, 86, 90, 87, 29, 32

**New Mean:**
\[
\text{Mean} = \frac{73 + 88 + 86 + 90 + 87 + 29 + 32}{7} = \frac{485}{7} \approx 69.29
\]

**New Standard Deviation:**
Let's recalculate the standard deviation with the new score added.

First, find the squared deviations from the new mean:
\[
\begin{align*}
(73 - 69.29)^2 &= 13.82 \\
(88 - 69.29)^2 &= 349.82 \\
(86 - 69.29)^2 &= 277.82 \\
(90 - 69.29)^2 &= 426.12 \\
(87 - 69.29)^2 &= 313.72 \\
(29 - 69.29)^2 &= 1621.56 \\
(32 - 69.29)^2 &= 1390.72 \\
\end{align*}
\]

Sum of squared deviations:
\[
\text{Sum} = 13.82 + 349.82 + 277.82 + 426.12 + 313.72 + 1621.56 + 1390.72 = 4393.58
\]

Variance:
\[
\text{Variance} = \frac{4393.58}{7} \approx 627.65
\]

Standard deviation:
\[
\text{Standard Deviation} = \sqrt{627.65} \approx 25.06
\]

### Summary
- **Initial Mean:** 75.5
- **Initial Standard Deviation:** 21.52
- **New Mean (after 7th quiz):** 69.29
- **New Standard Deviation (after 7th quiz):** 25.06

### Effect of the 7th Quiz Score
- **Mean:** The mean decreased from 75.5 to 69.29, reflecting the lower 7th score.
- **Standard Deviation:** The standard deviation increased from 21.52 to 25.06, indicating a greater spread in the scores due to the addition of the outlier score (32).

Do you have any questions or would you like further details?

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**Related Questions:**
1. What happens to the mean and standard deviation if Joe scores 95 on an 8th quiz?
2. How would the standard deviation change if Joe’s lowest score was removed?
3. What would the median of Joe’s scores be before and after the 7th quiz?
4. How does the presence of an outlier affect the mean and standard deviation?
5. Can you calculate the variance for the new set of 7 scores?

**Tip:** Standard deviation is sensitive to outliers, as seen with Joe’s low score of 29 and 32. This means a few very low or very high scores can significantly affect the spread of data.

Learn how to calculate the mean and standard deviation of a set of quiz scores, and understand how adding a new quiz score affects these statistics. Detailed step-by-step solution provided with specific data points. Suitable for high school students learning statistics.

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