Problem 26
Question
Two tomato plants were chosen from two groups of plants grown in different soils. Each sample plant produced 23 tomatoes The mean number of tomatoes for plants in the first soil was \(17,\) with a standard deviation of \(5 .\) The mean number of tomatoes for plants in the second soil was \(18,\) with a standard deviation of \(6 .\) Determine which plant, if either, is in the top 16\(\%\) of its group.
Step-by-Step Solution
Verified Answer
The plant from the first soil type is in the top 16% of its group.
1Step 1: Find the Z-score for each plant
Use the formula for the z-score, \(z = (X - \mu) / \sigma\), where X is the number of tomatoes, \mu is the mean and \sigma is the standard deviation. The z-score for the plant in the first soil type is \(z1 = (23 - 17) / 5 = 1.2\) and the z-score for the plant in the second soil type is \(z2 = (23 - 18) / 6 = 0.83.\)
2Step 2: Determine the percentile rank for each Z-score
Consult a Z-table to find the percentile rank for each Z-score. The percentile rank of the plant from the first soil type is approximately 88% and for the second soil type, it's approximately 80%.
3Step 3: Compare percentile ranks with threshold value
Compare each percentile rank with the threshold value of 84% (which corresponds to being top 16% as the percentile rank is a cumulative measure from the lower side). The plant from the first soil type that has a percentile rank of 88% is in the top 16% of its group, but the plant from the second soil type that has a percentile rank of 80% is not in the top 16% of its group.
Key Concepts
Z-scorepercentile rankstandard deviationmean
Z-score
A Z-score is a statistical tool that tells us how far a specific data point is from the mean of a data set. It is expressed in terms of standard deviations. To calculate the Z-score, we use the formula:
In our example, the Z-scores for the two tomato plants were 1.2 and 0.83. These scores indicate how many standard deviations the number of tomatoes produced by each plant is from the mean number of tomatoes produced by plants grown in their respective soils.
A Z-score can tell us whether a data point is typical for a dataset or if it stands out. In this case, positive Z-scores mean the plants produced more tomatoes than the average plant in their groups.
- \( z = \frac{X - \mu}{\sigma} \)
- \( X \) is the data point of interest,
- \( \mu \) is the mean of the data set, and
- \( \sigma \) is the standard deviation.
In our example, the Z-scores for the two tomato plants were 1.2 and 0.83. These scores indicate how many standard deviations the number of tomatoes produced by each plant is from the mean number of tomatoes produced by plants grown in their respective soils.
A Z-score can tell us whether a data point is typical for a dataset or if it stands out. In this case, positive Z-scores mean the plants produced more tomatoes than the average plant in their groups.
percentile rank
The percentile rank is a measure indicating the value below which a given percentage of observations in a group falls. If a plant is in the 88th percentile, it produced more tomatoes than 88% of the other plants in the same group.
We find the percentile rank using a Z-table. Once we determine the Z-score, we check this score against the Z-table to find the corresponding percentile.
We find the percentile rank using a Z-table. Once we determine the Z-score, we check this score against the Z-table to find the corresponding percentile.
- The plant in the first soil had a Z-score of 1.2, translating to a percentile rank of about 88%.
- The second soil plant had a Z-score of 0.83 and a percentile rank of approximately 80%.
standard deviation
Standard deviation is a crucial statistic that shows how much variation or dispersion exists from the mean in a dataset. A smaller standard deviation indicates that the data points tend to be close to the mean. A larger standard deviation suggests that the data points are spread out over a wider range.
In the example:
In the example:
- The first soil group has a standard deviation of \( 5 \), indicating a moderate level of tomato production variability.
- The second soil group has a standard deviation of \( 6 \), showing slightly more variability in the number of tomatoes produced.
mean
The mean, often called the average, is the sum of all values divided by the number of values in a dataset. It is one of the most common measures of central tendency, giving us a benchmark to measure variation throughout the data.
In the tomato plants example:
In the tomato plants example:
- The mean number of tomatoes for the first soil group is \( 17 \).
- For the second soil group, the mean is \( 18 \).
Other exercises in this chapter
Problem 25
Another measure of variation is variance, which equals \(\sigma^{2} .\) Find the variance and the standard deviation of each data set. $$ 0.7 \mathrm{g} \quad 0
View solution Problem 25
Find the area under each curve for the domain \(0 \leq x \leq 1\) $$ y=4 x+2 $$
View solution Problem 26
Open-Ended Describe a binomial experiment that can be solved using the expression \(7^{\mathrm{C}}_{2}(0.6)^{2}(0.4)^{5} .\)
View solution Problem 26
Describe how you could find the scores at the 20 th and 60 th percentiles in a set of 80 scores.
View solution