The annual precipitation data of a city is normally distributed with mean and standard deviation as 1000 mm and 200 mm, respectively. The probability that the annual precipitation will be more than 1200 mm is
Precipitation Data Analysis
The question provides details about the annual precipitation data of a city. This data is stated to be normally distributed, which is a common assumption in many natural phenomena studies, including rainfall patterns. We are given the mean and standard deviation of this distribution, and our goal is to determine the probability that the annual precipitation will exceed a specific value.
Understanding Normal Distribution for Precipitation
A normal distribution, often visualized as a "bell curve," is a symmetric probability distribution where data points cluster around the mean. For this problem, the characteristics of the annual precipitation are:
- Mean ($\mu$): The average annual precipitation is given as 1000 mm. This is the central value around which the precipitation data is distributed.
- Standard Deviation ($\sigma$): The measure of spread or variability in the precipitation data is 200 mm. A larger standard deviation indicates more variability in the precipitation amounts.
We need to find the probability that the annual precipitation (let's call it \($X$\)) will be more than 1200 mm, i.e., \($P(X > 1200 \text{ mm}) $\).
Standardizing Precipitation Values (Z-Score)
To calculate probabilities for a normal distribution, we typically convert the raw data value (\($X$\)) into a standard score, known as the Z-score. The Z-score tells us how many standard deviations an element is from the mean. The formula for the Z-score is:
\[Z = \frac{X - \mu}{\sigma}\]
Let's apply this formula to our specific value of 1200 mm:
- Given precipitation value, \($X = 1200 \text{ mm}$\)
- Mean, \($\mu = 1000 \text{ mm}$\)
- Standard deviation, \($\sigma = 200 \text{ mm}$\)
Now, let's calculate the Z-score:
\[Z = \frac{1200 - 1000}{200}\]
\[Z = \frac{200}{200}\]
\[Z = 1\]
So, an annual precipitation of 1200 mm is exactly 1 standard deviation above the mean.
Calculating Probability of Precipitation Exceeding 1200 mm
Now that we have the Z-score, we need to find \($P(Z > 1) $\) for the standard normal distribution. We know that the total area under the standard normal curve is 1, and it is symmetric around the mean (Z=0).
- The area to the right of the mean (Z=0) is 0.5 (or 50%).
- We need to find the area between Z=0 and Z=1, which can be looked up in a standard normal distribution table (Z-table).
- From the Z-table, the probability \($P(0 < Z < 1) $\) is approximately 0.3413.
To find \($P(Z > 1) $\), we subtract the area from Z=0 to Z=1 from the total area to the right of Z=0:
\[P(Z > 1) = P(Z > 0) - P(0 < Z < 1)\]
\[P(Z > 1) = 0.5 - 0.3413\]
\[P(Z > 1) = 0.1587\]
Converting this probability to a percentage:
\[0.1587 \times 100\% = 15.87\%\]
Comparing Probability with Given Options
The calculated probability that the annual precipitation will be more than 1200 mm is approximately 15.87%. Now, let's compare this value with the provided options:
- Option 1: Less than 50% (\($< 50\%$\)). Our calculated value of 15.87% falls into this category.
- Option 2: 50%. This would mean precipitation above the mean has a 50% chance, which is true for precipitation above 1000mm, but not 1200mm.
- Option 3: 75%. This is much higher than our calculated probability.
- Option 4: 100 %. This implies it's certain, which is incorrect.
Based on our calculation, 15.87% is indeed less than 50%.
Key Concepts for Precipitation Probability
- Normal Distribution: A bell-shaped, symmetric curve commonly used to model natural phenomena like precipitation.
- Mean ($\mu$): The average value of the data set, representing the center of the distribution.
- Standard Deviation ($\sigma$): A measure of the dispersion or spread of the data points around the mean.
- Z-score: A standardized score that indicates how many standard deviations a data point is from the mean. It allows comparison of values from different normal distributions.
- Probability: The likelihood of an event occurring, represented by the area under the probability distribution curve.
| Parameter | Value |
|---|---|
| Mean ($\mu$) | 1000 mm |
| Standard Deviation ($\sigma$) | 200 mm |
| Target Precipitation ($X$) | 1200 mm |
| Calculated Z-score | 1 |
| Probability ($P(Z > 1)$) | 0.1587 or 15.87% |
The probability that the annual precipitation will be more than 1200 mm is 15.87%, which is less than 50%.
Important Questions from Continuous Distributions
Suppose X is a continuous random variable with probability density function
\(f(x)=\frac{1}{\pi} \frac{1}{1+(x+1)^2}\), -∞ < x < ∞.
Define
\(Y=\left\{\begin{array}{cc} \frac{X}{|X|}, & \text { if } X \neq 0 \\ 0, & \text { if } X=0 \end{array}\right.\)
Then which of the following statements are true?
Let X1, X2, ..., Xn be a random sample from an absolutely continuous distribution with the probability density function
\(f(x \mid \theta)=\left\{\begin{array}{cl} e^{\theta-x}, & \text { if } x \geq \theta \\ 0, & \text { if } x<\theta \end{array},\right.\)
where θ ∈ ℝ is unknown. Define \(\bar{X}=\frac{1}{n} \sum_{i=1}^n X_i\) and X(1) = min{X1, ..., Xn}. Then
which of the following statements are true?
Suppose that X is a continuous random variable with probability density function given by:
f(x) = \(\left\{ {\begin{array}{c} {\frac{x}{8},}&{x \in \left[ {0,2} \right)}\\ {\frac{1}{4},}&{x \in \left[ {2,4} \right)}\\ { - \frac{x}{8} + \frac{3}{4},}&{x \in \left[ {4,6} \right)} \end{array}}\right.\)
Find the mean of X.
The variable x takes a value between 0 and 10 with uniform probability distribution. The variable y takes a value between 0 and 20 with uniform probability distribution. The probability of the sum of variables (x + y) being greater than 20 is _________
Probability density function of a random variable X is given below
\(f\left( x \right) = \left\{ {\begin{array}{*{20}{c}} {0.25}&{if\;1 \le x \le 5}\\ 0&{otherwise} \end{array}} \right.\)
P (X ≤ 4) is
