Understanding General Normal Distributions and Standardisation

What is the General Normal Distribution?

The general normal distribution is a continuous probability distribution that is symmetrical around its mean. In the previous chapter, we discussed a type of normal distribution known as the standard normal distribution. The standard normal distribution is a special case of the normal distribution with a mean of $0$ and a standard deviation of $1$. This chapter focuses on normal distributions that are symmetrical around values other than $1$.

Problems involving normal distributions are solved by using the same methods outlined in the previous chapter - using z-score tables. However, when dealing with general normal distributions, we must find the z-score from a given mean and standard deviation. This is done through a method called standardisation.

What is Standardisation?

Standardisation involves converting a normal distribution with a mean $\mu$ and standard deviation $\sigma$ into a standard normal distribution (with mean $0$ and standard deviation $1$). This is done, so we can apply the process of using z-score tables to determine a measure of probability. The process of standardisation is defined by the z-score formula, which is given by:

This formula standardises z-score values, so they can be measured according to the same scale, being the standard normal distribution, with a mean of $0$ and a standard deviation of $1$. As we know from the previous chapter, a sore indicates how many standard deviations a data point $x$ is from the mean. It is important to note that:

Solving Problems Using Standardisation

As discussed in the previous chapter, to determine the probability of certain values or ranges of a standard normal distribution, we must find corresponding values using a z-score table and graph the given bell curve. However, when dealing with a general normal distribution (a normal distribution with a mean value other than $1$), we must apply the process of standardisation, as seen above, before calculating the probability.

Practice Question 1

A class of students sit a math test graded out of $100$ marks. A student wishes to asses his performance against that of the rest of his class. The mean of all the test results was $75$ and the standard deviation of the results was $10$. This student achieved a score of $85$. Determine the student's performance on the test, relative to the average performance of all students who took the test. Additionally, create a relevant graph to support your answer. Use the following z-score table to assist you with any calculations.

z	.0	.1	.2	.3	.4	.5	.6	.7	.8	.9
0.	0.5000	0.5398	0.5793	0.6179	0.6554	0.6915	0.7257	0.7580	0.7881	0.8159
1.	0.8413	0.8643	0.8849	0.9032	0.9192	0.9332	0.9452	0.9554	0.9641	0.9713
2.	0.9772	0.9821	0.9861	0.9893	0.9918	0.9938	0.9953	0.9965	0.9974	0.9981
3.	0.9987	0.9990	0.9993	0.9995	0.9997	0.9998	0.9998	0.9999	0.9999	1.0000

The first step to solving this problem is to apply the process of standardisation. From the question we know that $x=85,\mu=75$ and $\sigma=10$.

$$z=\frac{x-\mu}{\sigma}\\=\frac{85-75}{10}\\\frac{10}{10}\\=1$$

Now that we have standardised the data given, we can see that the z-score of this distribution is $1$. Thus, we can use the z-score table provided in the question to determine the probability. From the table we can see that:

$$P(Z\leq 1 )\approx 0.8413$$

This means that the student scored better than approximately $84.13\%$ of students in the class on the exam. We can graph this by shading the area under a bell-curve graph, between $-\infty$ and $1$.

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