Calculating Fruit Fly Abdomen Length Standard Deviation
This problem requires calculating the standard deviation (SD) for a given set of measurements of male fruit fly abdomen lengths. The standard deviation measures the dispersion or spread of the data points around the mean.
Data Provided
- Abdomen Lengths (mm): 1.9, 2.4, 2.1, 2.0, 2.2, 2.4, 1.7, 1.8, 2.0, 2.0, 2.3, 2.1, 1.6, 2.3, 2.2
- Sample Size ($n$): 15
Standard Deviation Formula
For a sample dataset, the formula for sample standard deviation ($s$) is:
$s = \sqrt{\frac{\sum_{i=1}^{n}(x_i - \bar{x})^2}{n-1}}$
where $x_i$ represents each individual data point, $\bar{x}$ is the sample mean, and $n$ is the total number of data points.
Calculation Steps
-
Calculate the Sample Mean ($\bar{x}$):
First, sum all the abdomen length measurements:
$\sum x_i = 1.9 + 2.4 + 2.1 + 2.0 + 2.2 + 2.4 + 1.7 + 1.8 + 2.0 + 2.0 + 2.3 + 2.1 + 1.6 + 2.3 + 2.2 = 31.0$ mm.
Next, divide the sum by the sample size ($n$) to find the mean:
$\bar{x} = \frac{\sum x_i}{n} = \frac{31.0}{15} \approx 2.0667$ mm. -
Calculate the Standard Deviation ($s$):
Using the sample data and the calculated mean, the standard deviation is determined by applying the formula. The calculation yields a standard deviation of approximately $0.25$ mm.
Result
The computed standard deviation for the abdomen lengths of the male fruit flies is approximately $0.25$ mm, which corresponds to Option B.
Important Questions from Variance
Consider a distribution with the following probability density function $$f(x) = \begin{cases} 0.5, & 0 < x < 2 \\ 0.0, & Otherwise \end{cases}$$ Given that the mean of the above probability distribution is 1, the variance (rounded off to two decimal places) is _______________.
- Let $X$ and $Y$ be two independent random variables. $X$ follows $Bernoulli(p = 0.3)$ distribution and $Y$ follows $Normal(\mu = 0, \sigma^2 = 100)$ distribution.
Which of the following options is the variance of $(2X - 1)Y$? - For a given data set $\{x_1, x_2, \ldots, x_n\}$, where $n = 100$, it is known that
$$ \frac{1}{2000} \sum_{i=1}^{n} \sum_{j=1}^{n} (x_i - x_j)^2 = 99 $$
Let us denote $\bar{x} = \frac{1}{n} \sum_{i=1}^{n} x_i$.
The value of $\frac{1}{99} \sum_{i=1}^{n} (x_i - \bar{x})^2$ is __________ . (Answer in integer) - A random variable $X$ has the sample space $\{0,1\}$. The probability $P(X = 0) = 1/4$ and $P(X = 1) = 3/4$.
What is the variance of the random variable?
Hint: $\text{Mean } (\mu) = \sum_{i=1}^{n} x_i p(x_i) ; \text{Variance } (\sigma^2) = \sum_{i=1}^{n} (x_i - \mu)^2 p(x_i)$ The unbiased sample variance for the set of numbers: $S = \{40,45,50,55,60\}$ is_____. (write answer with one decimal place)
