Problem 6
Question
The following data represent the frequency distribution of the numbers of days that it took a certain ointment to clear up a skin rash: $$ \begin{array}{cc} \hline \text { Number of Days } & \text { Frequency } \\ \hline 1 & 2 \\ 2 & 7 \\ 3 & 9 \\ 4 & 27 \\ 5 & 11 \\ 6 & 5 \\ \hline \end{array} $$ Calculate the sample mean and the sample variance.
Step-by-Step Solution
Verified Answer
The sample mean is 3.87 days and the sample variance is 1.41 days squared.
1Step 1: Identify Midpoints and Frequencies
The midpoints for each class (in this case, each "Number of Days" is itself the midpoint) are 1, 2, 3, 4, 5, and 6. The corresponding frequencies are 2, 7, 9, 27, 11, and 5.
2Step 2: Determine Total Number of Observations
Calculate the total number of observations by summing all frequencies: \[ ext{Total Observations} = 2 + 7 + 9 + 27 + 11 + 5 = 61 \]
3Step 3: Calculate the Mean
Find the mean by calculating the weighted sum of midpoints and dividing by the total number of observations:\[ \text{Mean} = \frac{(1 \cdot 2) + (2 \cdot 7) + (3 \cdot 9) + (4 \cdot 27) + (5 \cdot 11) + (6 \cdot 5)}{61} \] Simplifying, we have:\[ \text{Mean} = \frac{2 + 14 + 27 + 108 + 55 + 30}{61} = \frac{236}{61} = 3.87 \]
4Step 4: Calculate Each Midpoint's Deviation from the Mean
Find the deviation of each midpoint from the mean, \(3.87\), and square these deviations:\[ (1 - 3.87)^2, (2 - 3.87)^2, (3 - 3.87)^2, (4 - 3.87)^2, (5 - 3.87)^2, (6 - 3.87)^2 \] These deviations are: 8.23, 3.50, 0.76, 0.02, 1.26, 4.49, respectively.
5Step 5: Calculate Weighted Sum of Squared Deviations
Use the squared deviations and frequencies to calculate the weighted sum:\[\sum = 2(8.23) + 7(3.50) + 9(0.76) + 27(0.02) + 11(1.26) + 5(4.49) \]Calculating each:\[16.46 + 24.50 + 6.84 + 0.54 + 13.86 + 22.45 = 84.65\]
6Step 6: Calculate the Sample Variance
Finally, divide the sum of squared deviations by the total number of observations minus one:\[ \text{Sample Variance} = \frac{84.65}{61 - 1} = \frac{84.65}{60} = 1.41 \]
Key Concepts
Understanding Frequency DistributionCalculating the Sample MeanExploring Sample Variance
Understanding Frequency Distribution
Frequency distribution is a way to organize data to show how often each value occurs. In our skincare ointment example, the number of days it took to clear the skin rash is listed alongside how many patients experienced each specific duration. Here’s why it’s helpful:
- It provides a quick visual representation of the dataset, making patterns easier to spot.
- It helps in understanding the distribution, such as which outcomes are most or least common.
- It often helps in calculating other statistics, like the mean or variance.
Calculating the Sample Mean
The sample mean is a crucial concept in statistics, used to find the average of your data set. It gives you a central value that represents your data. To calculate the sample mean:
- Use the formula \( \bar{x} = \frac{\sum{(x_i \times f_i)}}{n} \) where \( x_i \) is the midpoint and \( f_i \) is the frequency.
- Sum up the products of each midpoint and its corresponding frequency.
- Divide this total by the total number of observations (the sum of all frequencies).
Exploring Sample Variance
Sample variance is a measure of how spread out your data points are around the mean. It tells you whether the data points tend to be close to that mean (low variance) or far away from it (high variance).To find the sample variance, follow these steps:
- Calculate each data point's deviation from the mean (difference between each midpoint and the mean).
- Square each of these deviations to make them positive.
- Multiply each squared deviation by its respective frequency to get a weighted value.
- Sum up all these weighted squares.
- Divide by the total number of observations minus one \( (n - 1) \), not by \( n \) because it's a sample variance.
Other exercises in this chapter
Problem 5
An urn contains five blue and six green balls. You take two balls out of the urn, one after the other, without replacement. Find the probability that the second
View solution Problem 5
The Muesli-Mix is a popular breakfast hangout near a campus. A typical breakfast there consists of one beverage, one bowl of cereal, and a piece of fruit. If yo
View solution Problem 6
Let \(X\) be standard normally distributed. Use Chebyshev's inequality to estimate (a) \(P(|X| \geq 1)\), (b) \(P(|X| \geq 2)\), and (c) \(P(|X| \geq 3)\). Comp
View solution Problem 6
An urn contains five green balls, two blue balls, and three red balls. You remove three balls at random without replacement. Let \(X\) denote the number of red
View solution