Problem 209 - Applied Statistics and Probability for Engineers

Question

Without an automated irrigation system, the height of plants two weeks after germination is normally distributed with a mean of 2.5 centimeters and a standard deviation of 0.5 centimeter. (a) What is the probability that a plant's height is greater than 2.25 centimeters? (b) What is the probability that a plant's height is between 2.0 and 3.0 centimeters?

Step-by-Step Solution

Answer

(a) 0.6915; (b) 0.6826.

Step 1: Identify Distribution Parameters

The problem states that the height of plants follows a normal distribution with a mean \( \mu = 2.5 \) cm and a standard deviation \( \sigma = 0.5 \) cm. This information will be used to calculate probabilities.

Step 2: Calculate Z-Score for Part (a)

For part (a), to find the probability that the height is greater than 2.25 cm, calculate the Z-score using the formula:\[ Z = \frac{X - \mu}{\sigma} \]where \( X = 2.25 \) cm, \( \mu = 2.5 \) cm, and \( \sigma = 0.5 \) cm. Substituting these values, we get:\[ Z = \frac{2.25 - 2.5}{0.5} = -0.5 \]

Step 3: Find Probability from Z-Table for Part (a)

Using the Z-score calculated (-0.5), look up the corresponding probability in the Z-table. The Z-table gives the probability that a value is less than a given Z-score. For \( Z = -0.5 \), the probability is approximately 0.3085. Therefore, the probability of a plant's height being greater than 2.25 cm is \( 1 - 0.3085 = 0.6915 \).

Step 4: Calculate Z-Scores for Part (b)

For part (b), calculate the Z-scores for 2.0 cm and 3.0 cm heights.For 2.0 cm:\[ Z = \frac{2.0 - 2.5}{0.5} = -1.0 \]For 3.0 cm:\[ Z = \frac{3.0 - 2.5}{0.5} = 1.0 \]

Step 5: Find Probability for Range from Z-Table for Part (b)

Look up both Z-scores in the Z-table.- For \( Z = -1.0 \), the probability is approximately 0.1587.- For \( Z = 1.0 \), the probability is approximately 0.8413.The probability of a plant's height being between 2.0 and 3.0 cm is given by the difference:\[ 0.8413 - 0.1587 = 0.6826 \]

Step 6: Summarize the Solution

For part (a), the probability of height greater than 2.25 cm is 0.6915. For part (b), the probability of height between 2.0 and 3.0 cm is 0.6826.

Key Concepts

Z-score

The concept of a Z-score is a way to measure how far away a particular score is from the mean average of a dataset, expressed in terms of standard deviations. In probability and statistics, it is used to indicate how many standard deviations an element is from the mean of its distribution. When we calculate a Z-score, we subtract the mean from the data point of interest and then divide by the standard deviation. This results in a standard score that allows us to understand the position of a specific data point within the distribution.

• Formula: \( Z = \frac{X - \mu}{\sigma} \)
• \( X \) is the value or score.
• \( \mu \) is the mean of the data.
• \( \sigma \) is the standard deviation of the data.

Using a Z-table, we can then find the probability of a score occurring below, above, or between certain values.

Probability Calculation

Calculating probabilities using the normal distribution involves using the Z-score and Z-table. Once the Z-score is determined, we can refer to a Z-table, which shows the probability that a statistic is within a certain number of standard deviations from the mean. In the original exercise, we calculated two different probabilities: - The probability that a plant's height is greater than 2.25 cm was found by calculating the Z-score and identifying its probability from the table. This probability is the area to the right of the computed Z-score. - The probability of a height being between two values, such as between 2.0 cm and 3.0 cm, involves calculating the Z-score for both values. The difference between these probabilities gives the chance of the value falling in that range.

Standard Deviation

Standard deviation is a measure of the amount of variation or dispersion in a set of values. A low standard deviation indicates that values tend to be close to the mean of the set, while a high standard deviation indicates that the values are spread out over a wider range.

• It tells us how much the individual values of a dataset deviate from the mean.
• In any normal distribution, the standard deviation helps in determining the spread of the data.

In our exercise, the standard deviation was given as 0.5 cm. This value measures how much the heights of plants vary from the average height.

Mean

The mean, often referred to as the average, is the sum of all the numbers in a dataset divided by the number of values in that dataset. It is a measure of central tendency and gives us an overall idea of where the center of a dataset is located.

• Formula: \( \mu = \frac{\sum X}{N} \)
• \( \sum X \) is the sum of all values.
• \( N \) is the number of values.

In statistical problems involving normal distribution, the mean is a critical component. It forms the basis around which we calculate the probability and the Z-scores. In the context of the plant height example, the mean height of the plants was given as 2.5 cm. It tells us that, on average, this is the typical height you can expect the plants to reach two weeks after germination.