Frequency Consider the grouped sample data to the right. a) Calculate the approximate mean of these sample data. b) Calculate the approximate standard deviation of these sample data. Values 10 to under 30 to under 50 50 to under 70 70 to under 90 35 51 37 a) The approximate mean of these sample data is (Round to three decimal places as needed.) b) The approximate standard deviation of these sample data is (Round to three decimal places as needed.)The Correct Answer and Explanation is :
To calculate the approximate mean and standard deviation for grouped data, we need to use the following steps:
Data Summary:
The frequency table is as follows:
| Class Interval | Frequency (f) | Midpoint (x) |
|---|---|---|
| 10 to under 30 | 35 | 20 |
| 30 to under 50 | 51 | 40 |
| 50 to under 70 | 37 | 60 |
Step-by-step solution:
a) Calculate the approximate mean:
- Find the midpoint of each class:
- For the class interval 10 to under 30, the midpoint is ( (10 + 30)/2 = 20 )
- For the class interval 30 to under 50, the midpoint is ( (30 + 50)/2 = 40 )
- For the class interval 50 to under 70, the midpoint is ( (50 + 70)/2 = 60 )
- Multiply each frequency by its corresponding midpoint:
- ( 35 \times 20 = 700 )
- ( 51 \times 40 = 2040 )
- ( 37 \times 60 = 2220 )
- Find the sum of these products:
[
\text{Sum of (f \times x)} = 700 + 2040 + 2220 = 4960
] - Find the sum of the frequencies:
[
\text{Sum of f} = 35 + 51 + 37 = 123
] - Calculate the approximate mean:
[
\text{Mean} = \frac{\text{Sum of (f \times x)}}{\text{Sum of f}} = \frac{4960}{123} \approx 40.323
]
So, the approximate mean is ( \boxed{40.323} ).
b) Calculate the approximate standard deviation:
- Find the squared deviation of each midpoint from the mean, multiplied by the frequency:
[
(x – \mu)^2 = (20 – 40.323)^2 \approx 413.507
]
[
(x – \mu)^2 = (40 – 40.323)^2 \approx 0.105
]
[
(x – \mu)^2 = (60 – 40.323)^2 \approx 386.805
] - Multiply each squared deviation by the frequency:
- ( 35 \times 413.507 = 14472.745 )
- ( 51 \times 0.105 = 5.355 )
- ( 37 \times 386.805 = 14311.685 )
- Find the sum of these values:
[
\text{Sum of f} \times (x – \mu)^2 = 14472.745 + 5.355 + 14311.685 = 28889.785
] - Calculate the variance:
[
\text{Variance} = \frac{\text{Sum of (f \times (x – \mu)^2)}}{\text{Sum of f}} = \frac{28889.785}{123} \approx 234.187
] - Find the standard deviation:
[
\text{Standard deviation} = \sqrt{234.187} \approx 15.300
]
So, the approximate standard deviation is ( \boxed{15.300} ).
Explanation:
- Mean: The mean represents the center of the data, calculated by finding the weighted average of all class midpoints based on their frequencies.
- Standard deviation: The standard deviation measures the spread of the data. It is derived from the variance, which calculates how much each data point (or midpoint in this case) deviates from the mean. The larger the standard deviation, the more spread out the data.
I will now generate the image to illustrate this process.
Here is an infographic that illustrates the process of calculating the approximate mean and standard deviation for the grouped data. It visually walks through the steps and formulas involved in the calculation.
Let me know if you’d like further clarification or adjustments!
