Gaussian distribution probability

Question

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.

$$ f(x,μ,σ)= \large f(x,\mu , \sigma )=\frac{1}{\sigma \sqrt{2\pi}}\; e^{\frac{-(x- \mu)^{2}}{2\sigma ^{2}}} $$

1. 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?
2. 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?

Answer 1

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.

Answer 2

for the first question, the probability will be zero because weight is a continuous variable, and integrating between itself will result in zero when subtracting. for the second question: z = (x - μ)/σ z₁ = (155 - 180)/34 = -0.7353 z₂ = (120 - 180)/34 = -1.7647 use a z score table to look up the prabability for each value of z you can find one online on http://www.z-table.com? you'll find p₁ = 0.2327 and p₂ = 0.0392 probability is equal to p1-p2 = 0.1935

Answer 3

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 somewhere between 120 and 155 pounds?

To find P of weight is between 120 and 155 given average weight of 180...

For weight = 120 P(120|180) = 120 / 180 = 0.6667

For weight = 155 P(155|180) = 155 / 180 = 0.8611

Therefore to find P of weight to be between 120 and 155 given 180, we simply do

P(155|180) - P(120|180) = 0.8611 - 0.6667

Therefore $P = 0.1944$ Answer!