Use The Following Frequency Distribution Of Sample Data

Use The Following Frequency Distribution Of Sample Datain Answering

Use the following frequency distribution (of SAMPLE DATA) in answering the questions: class 0-10 f=10; class 10-20 f=20; class 20-30 f=40; class 30-40 f=20; class 40-50 f=10. QUESTION 2 what is the midpoint of the 3rd class? 5 points QUESTION 3 What is the total number of observations in the frequency distribution? 5 points QUESTION 4 What is the relative frequency of the 3rd class? 5 points QUESTION 5 In the formula for the standard deviation in this problem do you use "N" or "n-1"? 5 points QUESTION 6 what is the value of the mean? 40 points QUESTION 7 what is the value of the standard deviation?

Paper For Above instruction

The provided frequency distribution presents a sample dataset segmented into classes with respective frequencies. Each question addresses fundamental statistical measures derived from this data, including class midpoints, total observations, relative frequencies, and measures of central tendency and dispersion such as the mean and standard deviation. This comprehensive analysis facilitates understanding of the dataset’s distribution and variability.

Question 2: What is the midpoint of the 3rd class?

The class intervals are given as 0-10, 10-20, 20-30, 30-40, and 40-50. The third class interval is 20-30.

The midpoint (also called the class mark) is calculated as:

Midpoint = (Lower limit + Upper limit) / 2

For the 20-30 class:

Midpoint = (20 + 30) / 2 = 50 / 2 = 25

Thus, the midpoint of the third class is 25.

Question 3: What is the total number of observations in the frequency distribution?

To calculate the total observations, sum all the class frequencies:

Total observations = 10 + 20 + 40 + 20 + 10 = 100

Therefore, the total number of observations (\(N\)) is 100.

Question 4: What is the relative frequency of the 3rd class?

The relative frequency is the proportion of observations within a class relative to the total:

Relative frequency = (Class frequency) / (Total observations)

For the third class:

Relative frequency = 40 / 100 = 0.40

Expressed as a percentage, this is 40%.

Question 5: In the formula for the standard deviation in this problem do you use "N" or "n-1"?

When calculating the sample standard deviation, the denominator used is \(n-1\) to correct for bias in the estimation of the population standard deviation, especially when data is a sample rather than an entire population.

In this case, since the data appears to be a sample (based on the phrasing "sample data"), the formula should use \(n-1\).

Question 6: What is the value of the mean?

To compute the mean, the weighted average of class midpoints is used.

First, calculate the sum of the products of each class midpoint and its frequency, then divide by the total number of observations:

\[

\text{Mean} = \frac{\sum (f_i \times x_i)}{N}

\]

Where \(f_i\) is the frequency and \(x_i\) is the class midpoint.

Calculations:

- For class 0-10: midpoint = 5; \(f=10\); contribution = 10 × 5 = 50

- For class 10-20: midpoint = 15; \(f=20\); contribution = 20 × 15 = 300

- For class 20-30: midpoint = 25; \(f=40\); contribution = 40 × 25 = 1000

- For class 30-40: midpoint = 35; \(f=20\); contribution = 20 × 35 = 700

- For class 40-50: midpoint = 45; \(f=10\); contribution = 10 × 45 = 450

Sum of contributions: 50 + 300 + 1000 + 700 + 450 = 2500

Divide by total observations \(N=100\):

Mean = 2500 / 100 = 25

Question 7: What is the value of the standard deviation?

Calculating the standard deviation involves computing the variance first. Since the data are grouped into classes, the formula for the variance of grouped data is used:

\[

s^2 = \frac{\sum f_i (x_i - \bar{x})^2}{n-1}

\]

Step 1: Calculate the squared deviations of class midpoints from the mean (25) and multiply by class frequencies.

| Class | Midpoint \(x_i\) | \(f_i\) | \(x_i - \bar{x}\) | \((x_i - \bar{x})^2\) | \(f_i \times (x_i - \bar{x})^2\) |

|-------|------------------|---------|-------------------|------------------------|------------------------------|

| 0-10 | 5 | 10 | 5 - 25 = -20 | 400 | 10 × 400 = 4000 |

| 10-20 | 15 | 20 | 15 - 25 = -10 | 100 | 20 × 100 = 2000 |

| 20-30 | 25 | 40 | 0 | 0 | 40 × 0 = 0 |

| 30-40 | 35 | 20 | 35 - 25 = 10 | 100 | 20 × 100 = 2000 |

| 40-50 | 45 | 10 | 45 - 25 = 20 | 400 | 10 × 400 = 4000 |

Sum of \(f_i \times (x_i - \bar{x})^2\): 4000 + 2000 + 0 + 2000 + 4000 = 12,000

Step 2: Divide by \(n-1 = 100-1=99\):

\[

s^2 = \frac{12,000}{99} \approx 121.21

\]

Step 3: Take the square root to find standard deviation:

\[

s \approx \sqrt{121.21} \approx 11.01

\]

Thus, the standard deviation of the data is approximately 11.01.

Summary:

- The class midpoint of the third class (20-30) is 25.

- The total number of observations is 100.

- The relative frequency of the third class is 40%.

- For standard deviation calculations from grouped data, use \(n-1\).

- The mean of the data is 25.

- The standard deviation is approximately 11.01.

References

  • Bluman, A. G. (2014). Elementary Statistics: A Step By Step Approach. McGraw-Hill Education.
  • Freund, J. E. (2010). Modern Elementary Statistics. Pearson.
  • Urdan, T. C. (2017). Statistics in Plain English. Routledge.
  • Devore, J. L. (2015). Probability and Statistics for Engineering and the Sciences. Cengage Learning.
  • Mendenhall, W., Sincich, T. (2016). Statistics for Engineering and the Sciences. CRC Press.
  • Bacoder, N. K. (2012). Introduction to Statistics. World Scientific Publishing.
  • Ott, R. L., & Longnecker, M. (2010). An Introduction to Statistical Methods and Data Analysis. Brooks/Cole.
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  • Walpole, R. E., Myers, R. H., Myers, S. L., & Ye, K. (2012). Probability & Statistics for Engineers & Scientists. Pearson.

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