How To Find the Probability of Normal Distribution Given Mean and Standard Deviation
You are out fishing, and you wish to know the likelihood of catching fishes of a certain length. To calculate the probability of normal distribution in this case, you need to take the help of the Z-table. The Z-table is used to find the likelihood of an event happening within a specified set of parameters.
Let us understand the concept better with the help of a few problems and solving them. To ensure that your fundamentals in probability and statistics are sound, it is vital to know about types of data with examples. Data can be qualitative and quantitative and is collected through observation. This data know-how shall help you learn the concepts more interactively and engagingly – in a visual simulation environment, vis-a-vis a typical classroom setup.
Example 1 – The Mean and the Standard Deviation are Given
A fishing competition is being held in a pond where the fish lengths have a normal distribution given mean and standard deviation.
- What is the chance of catching fishes with smaller dimensions or say 6 inches?
- What is the chance of catching fishes that are quite large, say, 26 inches?
- And, what is the chance of catching fishes with average dimensions, or say between 16 inches to 26 inches?
The way to resolve this problem is to sketch out a normal distribution for fish lengths. Chart the fish length in inches on the x-axis. So, in this case, the median or average is 16 inches.
In the first question, you need to find the probability of fishes being 6 inches or lower – so, you need to find p(X<6). In the second question, you need to find p(X>26), and in the last question, you have to find the probability that the length of the fishes is between 16 inches to 26 inches or p(16<X<26).
Now, the final step involves using the z-formula to calculate the probability. So –
- In question 1, using the z-formula, the value of x becomes -2.
- In question 2, the z-formula converts the value of x to 0.
- In question 3, the z-formula makes the value of x to +2.
Understanding the terms and terminologies
- Normal distribution is a situation where the data is spread around a central line or value. When you make a graph or draw it out, you should get a Bell Curve to represent the Normal Distribution. In this case, the mean is equal to the median which is equal to the mode.
Remember, in the case of a normal distribution, there is symmetry along a central line. So, you can expect 50% of the values to be placed less than the mean and 50% of values are placed at higher values than the median.
- Standard deviation determines the spread of the given numbers from the central value in a Normal Distribution. When represented on a graph, the central line is valued at ‘0’, the deviations are represented at +1, -1, +2, -2, and so on.
Example 2 – Learning to Calculate the Mean and the Standard Deviation
95% of students in a school, when measured for their heights, lie between 1.0 meters and 1.8 meters tall. The data is normally distributed. Calculate the mean and the standard deviation.
Before you proceed ahead to the actual question, you can use the given data to find the mean or the average or the central line of the normal distribution.
It is (1.0+ 1.8) / 2 = 1.4 metres.
Now, 80% of students are in this height range. Before we calculate the standard deviation, you need to know that the standard deviation generally is measured by this rule:
- 68% are within 1 standard deviation from the mean;
- 95% are within 2 standard deviations from the mean;
- 99.7% are within 3 standard deviations from the mean.
Coming back to calculating the answer to the problem mentioned here, use the rule mentioned above. 95%, as is evident from the rule above, is 2 standard deviations on either side of the mean.
The value of 1 standard deviation, therefore, in this problem = (the highest value – the smallest value)/4. This is equal to 0.20 meters.
Now charting the value on the Bell Curve graph,
- 68% is between 1.2 meters and 1.6 meters and
- 95% is between 1.0 meters and 1.8 meters.
So, what does all this mean for standard deviation?
- A value is ‘likely’ to be within 1 Standard Deviation, which is 68 cases out of 100 will be.
- A value is ‘very likely’ to be within 2 Standard Deviation, which is 95 cases out 100 will be.
- A value is ‘almost certainly’ to be within 3 Standard Deviation, which is 99.7 cases out of 100 will be.
What is the Z-Score?
The distance or the number of standard deviations is what is called the z-score. So, in example 1, -2, 0, and +2 were the z-scores. In example 2, it is the 1 and 2 Standard Deviations that denote the z-score.
In example 2, if one of the students has a height of 2 meters, then what is the z-score or the number of standard deviation?
So, because the mean height is 1.4 meters, and 1 standard deviation is 0.2 meters, then the number of standard deviations for the student with a height of 2 meters is 0.2 meters x 3. So, in this example, 3 denotes the Standard Score or the Z-score.
To calculate the z-score, you need to first subtract the mean from the height given. In this case, it is 2.0 – 1.4 = 0.6 metres from the mean. Now, calculate the number of standard deviations. To do so divide the difference by the standard deviation = 0.6 metres /0.2 metres = 3 standard deviations.
Example 3 – Mean and the Standard Deviation are Given
In your city, the lifespan of the population is a normal distribution with a mean of 75 years. The standard deviation is 6 years. What is the probability that any person, A., will live more than 78 years of age?
Solution: First things first – assume the variable age of the person to be ‘x’. We need to find a solution for P(x>78).
The first step is to calculate the z-score. The mean given for the distribution is 75 years, and the standard deviation is 6 years. Therefore,
(78-75)/6 = 3/6 = 0.5
The probability P(x> 78) = 1 – P(x <= 78). You can use the normal distribution table to find the value of P(x <= 78) or P (Z<= 0.5).
As per the table, the value of P (Z<= 0.5) = 0.6915.
Thus, P(x> 78) = 1 – P(x <= 78) = 1 – P (Z<= 0.5) = 1 – 0.6915 = 0.3085.
What does this mean? That the probability of any random person in the city to live beyond 78 years is 0.3085!
Example 4 – Mean and Standard Deviation Needs to be Calculated
In a maths test, the results of 11 students were charted out:
20, 15, 26, 32, 18, 28, 35, 14, 26, 22, 17
These marks are out of 50, and only 5 students have got 50% marks. The rest are all likely to fail. Now, the teacher decides not to fail everyone in the test except those students who have not been able to score at least 1 Standard Deviation from the mean value.
Solving this problem, you need to first calculate the mean, which is 23. Next, calculate the Standard Deviation, which is 6.6. Plotting the z-scores, you will find that only 2 students of the lot got lower than -1 Standard deviation (-1.21 and -1.35). Thus, the professor will fail these two students.
Understanding the z-score formula
The formula is: Z = (x- μ)/ σ
In this case,
z stands for z-score
μ or mu denotes the mean
σ or sigma stands for the standard deviation.
What is the Benefit of Standardizing?
The process of finding the Standard Score or the z-score is called Standardizing. There are multiple benefits of doing this, and this is a method that has practical applicability in our day-to-day lives.
- One of the biggest benefits is that it helps us analyze data and then accordingly make decisions.
- Interpreting random data can be tedious and complicated. However, in standardization, we use the Standard Normal Distribution Table to do so. Here, all figures are mentioned that make things easy and simple for individual calculations.
For getting accurate values of mean, standard deviation, and z-score, it is always advisable to use the Standard Normal Distribution Table. The table is readily available where the probabilities of different ranges of normal distribution have been depicted and can be referred to instantly without using integral calculus to determine the probabilities.
Conclusion
As is evident, the concept of calculating the probability of normal distribution given mean and standard deviation (or relevant data is given to calculate the mean and Standard Deviation) is interesting if and only if you understand the concepts, the rules, and the application. Clarity is very important – rest, it is easy and quite simple. Remember, the method, understand the problem, and then solving is not a big issue. To understand the concepts thoroughly, ensure that you get enrolled with the best maths tutors who will help you in a simulated environment.
