Answer to Question #155379 in Statistics and Probability for rock 206

Question

suppose the mean weight of all the students in this examination hall is 54 kgs, and the standard deviation is 4 kgs. what percentage of students will fall between 44 to 62 kgs? justify your answer

Answer 1

We assume that the student weight distribution can be well approximated by a normal distribution with the same parameters: mean 54 kgs and the standard deviation 4 kgs.

Then the student weight is equal to 54+4X, where X - a standard normal distributed random variable. The probability that the student weight is between 44 and 62 kgs is equal to the probability that the value of X is between -2.5 and 2:

$44\leq 54+4X\leq62$

$(44-54)/4\leq X\leq (62-54)/4$

$-2.5\leq X\leq 2$

The value of this probability may be found as the integral

$\frac{1}{\sqrt{2\pi}}\int\limits_{-2.5}^{2}e^{-x^2/2}dx = \Phi(2) - \Phi(-2.5) = 0.9772 - 0.0018 = 0.9754$

where $\Phi(x) = \frac{1}{\sqrt{2\pi}}\int\limits_{-\infty}^{x}e^{-t^2/2}dt$ - the cumulative distribution function of the standard normal distribution, its values we may get from the tables.

Answer. The probability that the student weight is between 44 and 62 kgs is equal to 97.54%