Question

1. given the histogram below, calculate the mean and standard deviation. label μ, σ, and - σ on the histogram. use these values to determine whether the distribution could be normal.
2. mean =
3. standard deviation =
4. μ - σ =
5. μ + σ =
6. what percent of the data is within one σ of μ?
7. does the distribution appear normal?

Answer:

To solve this problem, we first need to determine the data values from the histogram. Let's assume the midpoints of each bar (or the data values) and their frequencies:

• Let's list the data points:
• For the first bar (around 45): 2 data points (count = 2)
• Second bar (around 46): 1 data point (count = 1)
• Third bar (around 47): 4 data points (count = 4)
• Fourth bar (around 48): 5 data points (count = 5)
• Fifth bar (around 49): 5 data points (count = 5)
• Sixth bar (around 50): 2 data points (count = 2)
• Seventh bar (around 51): 3 data points (count = 3)

Now, let's calculate the mean, standard deviation, and other values.

Step 1: Calculate the Mean ($\mu$)

The mean is calculated as $\mu = \frac{\sum (x_i \cdot f_i)}{\sum f_i}$, where $x_i$ is the data point and $f_i$ is the frequency.

First, calculate $\sum (x_i \cdot f_i)$:

• $45 \cdot 2 = 90$
• $46 \cdot 1 = 46$
• $47 \cdot 4 = 188$
• $48 \cdot 5 = 240$
• $49 \cdot 5 = 245$
• $50 \cdot 2 = 100$
• $51 \cdot 3 = 153$

Sum of $x_i \cdot f_i$: $90 + 46 + 188 + 240 + 245 + 100 + 153 = 1062$

Sum of frequencies ($\sum f_i$): $2 + 1 + 4 + 5 + 5 + 2 + 3 = 22$

Mean: $\mu = \frac{1062}{22} \approx 48.27$ (Wait, this seems off. Maybe the midpoints are different. Let's re-examine the histogram. The x-axis is labeled 44, 46, 48, 50, 52. So the bins are 44-46, 46-48, 48-50, 50-52? Wait, the count for each bin:

• 44-46: count = 2 (midpoint 45)
• 46-48: count = 1 (midpoint 47)
• 48-50: count = 4 + 5 + 5 = 14? Wait, no, the histogram has bars at 45 (count 2), 46 (count 1), 47 (count 4), 48 (count 5), 49 (count 5), 50 (count 2), 51 (count 3). Maybe the x-axis is the data value, not the bin. Let's assume the data points are:

• 45: 2 times
• 46: 1 time
• 47: 4 times
• 48: 5 times
• 49: 5 times
• 50: 2 times
• 51: 3 times

Now, recalculate the mean:

$\sum (x_i \cdot f_i) = 45*2 + 46*1 + 47*4 + 48*5 + 49*5 + 50*2 + 51*3$

$= 90 + 46 + 188 + 240 + 245 + 100 + 153 = 1062$

$\sum f_i = 2 + 1 + 4 + 5 + 5 + 2 + 3 = 22$

$\mu = 1062 / 22 ≈ 48.27$ (This seems inconsistent with the x-axis labels. Maybe the x-axis is the data value, and the bars are at 45, 46, 47, 48, 49, 50, 51. Alternatively, maybe the data is:

• 45: 2
• 46: 1
• 47: 4
• 48: 5
• 49: 5
• 50: 2
• 51: 3

Now, let's calculate the standard deviation.

Step 2: Calculate the Standard Deviation ($\sigma$)

The standard deviation is calculated as $\sigma = \sqrt{\frac{\sum f_i (x_i - \mu)^2}{\sum f_i}}$

First, calculate $(x_i - \mu)^2$ for each $x_i$:

• For $x = 45$: $(45 - 48.27)^2 ≈ 10.69$; $f_i = 2$; $f_i (x_i - \mu)^2 ≈ 2 * 10.69 ≈ 21.38$
• For $x = 46$: $(46 - 48.27)^2 ≈ 5.15$; $f_i = 1$; $f_i (x_i - \mu)^2 ≈ 5.15$
• For $x = 47$: $(47 - 48.27)^2 ≈ 1.61$; $f_i = 4$; $f_i (x_i - \mu)^2 ≈ 4 * 1.61 ≈ 6.44$
• For $x = 48$: $(48 - 48.27)^2 ≈ 0.07$; $f_i = 5$; $f_i (x_i - \mu)^2 ≈ 5 * 0.07 ≈ 0.35$
• For $x = 49$: $(49 - 48.27)^2 ≈ 0.53$; $f_i = 5$; $f_i (x_i - \mu)^2 ≈ 5 * 0.53 ≈ 2.65$
• For $x = 50$: $(50 - 48.27)^2 ≈ 2.99$; $f_i = 2$; $f_i (x_i - \mu)^2 ≈ 2 * 2.99 ≈ 5.98$
• For $x = 51$: $(51 - 48.27)^2 ≈ 7.45$; $f_i = 3$; $f_i (x_i - \mu)^2 ≈ 3 * 7.45 ≈ 22.35$

Now, sum these values: $21.38 + 5.15 + 6.44 + 0.35 + 2.65 + 5.98 + 22.35 ≈ 64.3$

Then, $\sigma = \sqrt{\frac{64.3}{22}} ≈ \sqrt{2.92} ≈ 1.71$

Step 3: Calculate $\mu - \sigma$ and $\mu + \sigma$

$\mu - \sigma ≈ 48.27 - 1.71 ≈ 46.56$

$\mu + \sigma ≈ 48.27 + 1.71 ≈ 49.98$

Step 4: Determine the Percentage of Data Within $\mu \pm \sigma$

We need to count the number of data points between 46.56 and 49.98.

Data points:

• 45: 2 (below 46.56)
• 46: 1 (below 46.56)
• 47: 4 (between 46.56 and 49.98)
• 48: 5 (between 46.56 and 49.98)
• 49: 5 (between 46.56 and 49.98)
• 50: 2 (above 49.98)
• 51: 3 (above 49.98)

Number of data points within the range: $4 + 5 + 5 = 14$

Total data points: 22

Percentage: $\frac{14}{22} \approx 63.64\%$

Step 5: Determine if the Distribution is Normal

A normal distribution should have approximately 68% of the data within $\mu \pm \sigma$. Our calculated percentage is ~63.64%, which is close to 68%, but not exact. Additionally, the histogram should be symmetric. Let's check the frequencies:

• Left of mean (48.27): 2 (45) + 1 (46) + 4 (47) + 5 (48) = 12
• Right of mean: 5 (49) + 2 (50) + 3 (51) = 10

The distribution is somewhat symmetric but not perfectly. However, the percentage within $\mu \pm \sigma$ is close to 68%, so it could be approximately normal.

Final Answers:

1. mean = $\boldsymbol{48.27}$ (approximate)
2. standard deviation = $\boldsymbol{1.71}$ (approximate)
3. $\mu - \sigma$ = $\boldsymbol{46.56}$ (approximate)
4. $\mu + \sigma$ = $\boldsymbol{49.98}$ (approximate)
5. What percent of the data is within one $\sigma$ of $\mu$? $\boldsymbol{63.64\%}$ (approximate)
6. Does the distribution appear Normal? $\boldsymbol{Yes}$ (it is approximately normal, close to 68% within $\mu \pm \sigma$ and somewhat symmetric)

(Note: The exact values may vary depending on the precise data points and their frequencies. The above calculations are based on the assumed data points from the histogram.)