Let's tackle the problem step by step.
a. Finding the mean (average) weight of the mice:
To find the mean, we need to use the following formula for a frequency distribution:
[tex]\[ \text{Mean} = \frac{\sum (f_i \cdot x_i)}{N} \][/tex]
where:
- [tex]\( f_i \)[/tex] is the frequency of the [tex]\(i\)[/tex]-th interval,
- [tex]\( x_i \)[/tex] is the midpoint of the [tex]\(i\)[/tex]-th interval, and
- [tex]\( N \)[/tex] is the total number of observations.
Given intervals and frequencies, first we find the midpoints of each interval:
- For [tex]\(41.5-43.5\)[/tex], midpoint [tex]\(x_1 = \frac{41.5 + 43.5}{2} = 42.5\)[/tex]
- For [tex]\(43.5-45.5\)[/tex], midpoint [tex]\(x_2 = \frac{43.5 + 45.5}{2} = 44.5\)[/tex]
- For [tex]\(45.5-47.5\)[/tex], midpoint [tex]\(x_3 = \frac{45.5 + 47.5}{2} = 46.5\)[/tex]
- For [tex]\(47.5-49.5\)[/tex], midpoint [tex]\(x_4 = \frac{47.5 + 49.5}{2} = 48.5\)[/tex]
- For [tex]\(49.5-51.5\)[/tex], midpoint [tex]\(x_5 = \frac{49.5 + 51.5}{2} = 50.5\)[/tex]
- For [tex]\(51.5-53.5\)[/tex], midpoint [tex]\(x_6 = \frac{51.5 + 53.5}{2} = 52.5\)[/tex]
- For [tex]\(53.5-55.5\)[/tex], midpoint [tex]\(x_7 = \frac{53.5 + 55.5}{2} = 54.5\)[/tex]
- For [tex]\(55.5-57.5\)[/tex], midpoint [tex]\(x_8 = \frac{55.5 + 57.5}{2} = 56.5\)[/tex]
- For [tex]\(57.5-59.5\)[/tex], midpoint [tex]\(x_9 = \frac{57.5 + 59.5}{2} = 58.5\)[/tex]
Next, we use the given frequencies [tex]\( [2, 7, 11, 24, 16, 16, 15, 7, 2] \)[/tex] and the midpoints to calculate the mean:
[tex]\[
\text{Mean} = \frac{(2 \cdot 42.5) + (7 \cdot 44.5) + (11 \cdot 46.5) + (24 \cdot 48.5) + (16 \cdot 50.5) + (16 \cdot 52.5) + (15 \cdot 54.5) + (7 \cdot 56.5) + (2 \cdot 58.5)}{100}
\][/tex]
[tex]\[
\text{Mean} = \frac{85 + 311.5 + 511.5 + 1164 + 808 + 840 + 817.5 + 395.5 + 117}{100}
\][/tex]
[tex]\[
\text{Mean} = \frac{5050}{100} = 50.5
\][/tex]
So, the mean weight of the mice is 50.50 grams.
b. Finding the standard deviation of the weight of the mice:
To find the standard deviation, we first calculate the variance using the formula:
[tex]\[ \text{Variance} = \frac{\sum f_i (x_i - \text{mean})^2}{N} \][/tex]
Then, the standard deviation is the square root of the variance.
Using the previously calculated mean ([tex]\( \text{mean} = 50.5 \)[/tex]):
[tex]\[
\text{Variance} = \frac{2(42.5 - 50.5)^2 + 7(44.5 - 50.5)^2 + 11(46.5 - 50.5)^2 + 24(48.5 - 50.5)^2 + 16(50.5 - 50.5)^2 + 16(52.5 - 50.5)^2 + 15(54.5 - 50.5)^2 + 7(56.5 - 50.5)^2 + 2(58.5 - 50.5)^2}{100}
\][/tex]
[tex]\[
\text{Variance} = \frac{2 \cdot 64 + 7 \cdot 36 + 11 \cdot 16 + 24 \cdot 4 + 16 \cdot 0 + 16 \cdot 4 + 15 \cdot 16 + 7 \cdot 36 + 2 \cdot 64}{100}
\][/tex]
[tex]\[
\text{Variance} = \frac{128 + 252 + 176 + 96 + 0 + 64 + 240 + 252 + 128}{100} = \frac{1336}{100} = 13.36
\][/tex]
Now, calculating the standard deviation:
[tex]\[
\text{Standard Deviation} = \sqrt{13.36} \approx 3.66
\][/tex]
So, the standard deviation of the weight of the mice is 3.66 grams.
