Math / Data & Statistics

Question) The following distribution shows the exam scores for all students in a statistics class. a. Complete the table above. b. Draw a histogram. c. Find the mean with one decimal place. d. Find the standard derivation with one decimal place. e. Interpret the shape of the distribution. (Hint: skewed to the left, skewed to the right, or

Studdy Solution

STEP 1

1. The table is partially filled with class boundaries, frequencies, and some calculations.
2. The total frequency is given as 19, and the total for $f.x$ is 1843.0.
3. The mean and standard deviation need to be calculated to one decimal place.
4. The histogram will be based on the class boundaries and frequencies.
5. The interpretation of the distribution shape will be based on the histogram and calculated statistics.

STEP 2

1. Complete the table.
2. Draw a histogram.
3. Calculate the mean.
4. Calculate the standard deviation.
5. Interpret the shape of the distribution.

STEP 3

Complete the table by calculating the missing values. This involves: - Calculating the midpoints for each class. - Calculating $f.x$ for each class. - Calculating $x - \bar{x}$ for each class. - Calculating $(x - \bar{x})^2$ for each class. - Calculating $f(x - \bar{x})^2$ for each class.

STEP 4

Calculate the midpoint for each class boundary. For example, for the class boundary 47.5-56.5, the midpoint is:
$\text{Midpoint} = \frac{47.5 + 56.5}{2} = 52.0$
Repeat for all class boundaries.

STEP 5

Calculate $f.x$ for each class by multiplying the frequency by the midpoint. For example, if the frequency for the class 47.5-56.5 is $f$ , then:
$f.x = f \times 52.0$
Repeat for all classes.

STEP 6

Calculate the mean $\bar{x}$ using the formula:
$\bar{x} = \frac{\sum f.x}{\sum f} = \frac{1843.0}{19}$
Calculate $\bar{x}$ .

STEP 7

Draw a histogram using the class boundaries and frequencies. Each bar represents a class interval, with the height corresponding to the frequency.

STEP 8

Calculate $x - \bar{x}$ for each class using the midpoints and the mean calculated in STEP_4.

STEP 9

Calculate $(x - \bar{x})^2$ for each class.

STEP 10

Calculate $f(x - \bar{x})^2$ for each class by multiplying the frequency by $(x - \bar{x})^2$ .

STEP 11

Calculate the standard deviation using the formula:
$\sigma = \sqrt{\frac{\sum f(x - \bar{x})^2}{\sum f}}$
Calculate $\sigma$ to one decimal place.

STEP 12

Interpret the shape of the distribution based on the histogram and calculated statistics. Determine if it is skewed to the left, skewed to the right, or symmetric.
The mean is calculated as:
$\bar{x} = \frac{1843.0}{19} = 97.0$
The standard deviation is calculated as:
$\sigma = \sqrt{\frac{\sum f(x - \bar{x})^2}{19}}$
Interpretation of the distribution shape will depend on the skewness observed in the histogram and the relationship between the mean and median.

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Studdy Solution

STEP 1

STEP 2

1. Complete the table.
2. Draw a histogram.
3. Calculate the mean.
4. Calculate the standard deviation.
5. Interpret the shape of the distribution.

STEP 3

Complete the table by calculating the missing values. This involves: - Calculating the midpoints for each class. - Calculating $f.x$ for each class. - Calculating $x - \bar{x}$ for each class. - Calculating $(x - \bar{x})^2$ for each class. - Calculating $f(x - \bar{x})^2$ for each class.

STEP 4

Calculate the midpoint for each class boundary. For example, for the class boundary 47.5-56.5, the midpoint is:
$\text{Midpoint} = \frac{47.5 + 56.5}{2} = 52.0$
Repeat for all class boundaries.

STEP 5

Calculate $f.x$ for each class by multiplying the frequency by the midpoint. For example, if the frequency for the class 47.5-56.5 is $f$ , then:
$f.x = f \times 52.0$
Repeat for all classes.

STEP 6

Calculate the mean $\bar{x}$ using the formula:
$\bar{x} = \frac{\sum f.x}{\sum f} = \frac{1843.0}{19}$
Calculate $\bar{x}$ .

STEP 7

Draw a histogram using the class boundaries and frequencies. Each bar represents a class interval, with the height corresponding to the frequency.

STEP 8

Calculate $x - \bar{x}$ for each class using the midpoints and the mean calculated in STEP_4.

STEP 9

Calculate $(x - \bar{x})^2$ for each class.

STEP 10

Calculate $f(x - \bar{x})^2$ for each class by multiplying the frequency by $(x - \bar{x})^2$ .

STEP 11

Calculate the standard deviation using the formula:
$\sigma = \sqrt{\frac{\sum f(x - \bar{x})^2}{\sum f}}$
Calculate $\sigma$ to one decimal place.

STEP 12

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Studdy Solution

STEP 1

STEP 2

1. Complete the table.2. Draw a histogram.3. Calculate the mean.4. Calculate the standard deviation.5. Interpret the shape of the distribution.

STEP 3

STEP 4

Calculate the midpoint for each class boundary. For example, for the class boundary 47.5-56.5, the midpoint is:Midpoint=47.5+56.52=52.0 \text{Midpoint} = \frac{47.5 + 56.5}{2} = 52.0 Midpoint=247.5+56.5​=52.0Repeat for all class boundaries.

STEP 5

Calculate f.x f.x f.x for each class by multiplying the frequency by the midpoint. For example, if the frequency for the class 47.5-56.5 is f f f, then:f.x=f×52.0 f.x = f \times 52.0 f.x=f×52.0Repeat for all classes.

STEP 6

Calculate the mean xˉ \bar{x} xˉ using the formula:xˉ=∑f.x∑f=1843.019 \bar{x} = \frac{\sum f.x}{\sum f} = \frac{1843.0}{19} xˉ=∑f∑f.x​=191843.0​Calculate xˉ \bar{x} xˉ.

STEP 7

Draw a histogram using the class boundaries and frequencies. Each bar represents a class interval, with the height corresponding to the frequency.

STEP 8

Calculate x−xˉ x - \bar{x} x−xˉ for each class using the midpoints and the mean calculated in STEP_4.

STEP 9

Calculate (x−xˉ)2 (x - \bar{x})^2 (x−xˉ)2 for each class.

STEP 10

Calculate f(x−xˉ)2 f(x - \bar{x})^2 f(x−xˉ)2 for each class by multiplying the frequency by (x−xˉ)2 (x - \bar{x})^2 (x−xˉ)2.

STEP 11

Calculate the standard deviation using the formula:σ=∑f(x−xˉ)2∑f \sigma = \sqrt{\frac{\sum f(x - \bar{x})^2}{\sum f}} σ=∑f∑f(x−xˉ)2​​Calculate σ \sigma σ to one decimal place.

STEP 12

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1. Complete the table.
2. Draw a histogram.
3. Calculate the mean.
4. Calculate the standard deviation.
5. Interpret the shape of the distribution.

Calculate the midpoint for each class boundary. For example, for the class boundary 47.5-56.5, the midpoint is:
$\text{Midpoint} = \frac{47.5 + 56.5}{2} = 52.0$
Repeat for all class boundaries.

Calculate $f.x$ for each class by multiplying the frequency by the midpoint. For example, if the frequency for the class 47.5-56.5 is $f$ , then:
$f.x = f \times 52.0$
Repeat for all classes.

Calculate the mean $\bar{x}$ using the formula:
$\bar{x} = \frac{\sum f.x}{\sum f} = \frac{1843.0}{19}$
Calculate $\bar{x}$ .

Calculate $x - \bar{x}$ for each class using the midpoints and the mean calculated in STEP_4.

Calculate $(x - \bar{x})^2$ for each class.

Calculate $f(x - \bar{x})^2$ for each class by multiplying the frequency by $(x - \bar{x})^2$ .

Calculate the standard deviation using the formula:
$\sigma = \sqrt{\frac{\sum f(x - \bar{x})^2}{\sum f}}$
Calculate $\sigma$ to one decimal place.