MATH SOLVE

2 months ago

Q:
# For a normal distribution with a mean of 16 and a standard deviation of 3, find the 50th percentile. Can someone please explain how to solve this.

Accepted Solution

A:

The probability distribution of normal distribution is given by,

f(x) = [tex] \frac{1}{{ \sqrt{2 \pi \alpha ^{2} } }} e^{ \frac{-(x-a)^{2} }{2 \alpha^{2} } } [/tex]

Where a = mean = 16 and α = standard deviation = 3

In normal distribution, skewness is zero. That is, distribution is symmetric.

Therefore, mean, median and mode coincide at a point in case of normal distribution.

50th percentile is same as median or mean or mode

Therefore, 50th percentile = 16

f(x) = [tex] \frac{1}{{ \sqrt{2 \pi \alpha ^{2} } }} e^{ \frac{-(x-a)^{2} }{2 \alpha^{2} } } [/tex]

Where a = mean = 16 and α = standard deviation = 3

In normal distribution, skewness is zero. That is, distribution is symmetric.

Therefore, mean, median and mode coincide at a point in case of normal distribution.

50th percentile is same as median or mean or mode

Therefore, 50th percentile = 16