I am using Gaussian distribution to solve the below question. By applying
below formula, I could find the answer to the first question as 0.
Does this mean that the probability of getting 185 pounds man is zero?
Also How to solve the second question using this formula and explain when to use Gaussian distribution and its meaning and how different it is from Binomial Distribution.
$$
f(x,μ,σ)= \large f(x,\mu , \sigma )=\frac{1}{\sigma \sqrt{2\pi}}\; e^{\frac{-(x- \mu)^{2}}{2\sigma ^{2}}}
$$
- Assume the average weight of an American adult male is 180 pounds
with a standard deviation of 34 pounds. The distribution of weights
follows a normal distribution. What is the probability that a man
weighs exactly 185 pounds? - Like in the previous question, assume the average weight of an
American adult male is 180 pounds with a standard deviation of 34
pounds. The distribution of weights follows a normal distribution.
What is the probability that a man weighs somewhere between 120 and
155 pounds?
Best Answer
Clearly, you need to brush up on what a PDF (Probability Density Function) means. $P(X = 185) = 0$ but not in the way that you think. Probabilities from a PDF are calculated through an integral. Area under the curve gives us what the probability is for $x$ to be in that range.
Hence , $P(X = 185) = \int_{185}^{185}f(x,\mu, \sigma)dx = 0$.
Similarly, you can find the $P(120<X<155) = \int_{120}^{155}f(x,\mu, \sigma)dx$
One more thing to note is that you cannot calculate this integral directly. You will need online calculators to find the answer to that.