How can you estimate the parameters of a normal distribution?
Answer
614.7k+ views
Hint: In this question we will first write the general probability density function abbreviated as pdf for the normal function and then look at all the parameters which are required to get the probability. We will then see a random sample and then derive the formulas which are required to get the parameters of a normal distribution.
Complete step-by-step answer:
The normal distribution has a probability density function as:
$f\left( x \right)=\dfrac{1}{\sqrt{2\pi {{\sigma }^{2}}}}{{e}^{-\dfrac{\left( x-\mu \right)}{2{{\sigma }^{2}}}}}$, where $\mu $is the mean of the distribution and $\sigma $ is the standard deviation of the distribution.
Now in the probability density function we have $\pi $ and $e$ which are constant values and not variables.
Since there are no other variables, the mean $\mu $ and the standard deviation $\sigma $ are the two parameters which are required to get the value of $f\left( x \right)$.
Now when we have a random sample with elements, the mean can be calculated as:
$\mu =\dfrac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}+.....+{{x}_{n}}}{n}$
And the standard deviation of the sample can be calculated as:
$\sigma =\sqrt{\dfrac{{{\left( {{x}_{1}}-\mu \right)}^{2}}+{{\left( {{x}_{2}}-\mu \right)}^{2}}+{{\left( {{x}_{3}}-\mu \right)}^{2}}+....+{{\left( {{x}_{n}}-\mu \right)}^{2}}}{n}}$
Now on calculating these two values they can be substituted in the probability density function of the normal distribution to get the required solution.
Note: It is to be remembered that the mean of the sample is also called as the average of the sample. Standard deviation represents how much the data varies from each other in a given sample, the less the variance implies that the sample is more uniformly spread. There also exist other types of probability density functions like the Bernoulli distribution, Poisson distribution, Binomial distribution etc.
Complete step-by-step answer:
The normal distribution has a probability density function as:
$f\left( x \right)=\dfrac{1}{\sqrt{2\pi {{\sigma }^{2}}}}{{e}^{-\dfrac{\left( x-\mu \right)}{2{{\sigma }^{2}}}}}$, where $\mu $is the mean of the distribution and $\sigma $ is the standard deviation of the distribution.
Now in the probability density function we have $\pi $ and $e$ which are constant values and not variables.
Since there are no other variables, the mean $\mu $ and the standard deviation $\sigma $ are the two parameters which are required to get the value of $f\left( x \right)$.
Now when we have a random sample with elements, the mean can be calculated as:
$\mu =\dfrac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}+.....+{{x}_{n}}}{n}$
And the standard deviation of the sample can be calculated as:
$\sigma =\sqrt{\dfrac{{{\left( {{x}_{1}}-\mu \right)}^{2}}+{{\left( {{x}_{2}}-\mu \right)}^{2}}+{{\left( {{x}_{3}}-\mu \right)}^{2}}+....+{{\left( {{x}_{n}}-\mu \right)}^{2}}}{n}}$
Now on calculating these two values they can be substituted in the probability density function of the normal distribution to get the required solution.
Note: It is to be remembered that the mean of the sample is also called as the average of the sample. Standard deviation represents how much the data varies from each other in a given sample, the less the variance implies that the sample is more uniformly spread. There also exist other types of probability density functions like the Bernoulli distribution, Poisson distribution, Binomial distribution etc.
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