debbie_diaz
New member
- Joined
- Feb 11, 2022
- Messages
- 1
Given a distribution with pdf f(x), the median is the number m such that
m
(integral)
−∞. f(x) dx= 0.5.
That is, if X is a random variable with this distribution, then P(X≤m) = 0.5.
The median is also known as the 50th percentile. We can also consider other
percentiles. For example, if we want the 23rd percentile, then we try to find a
number q such that
q
(integral)
−∞ f(x) dx= 0.23.(1)
The general notation used by the textbook is that for any number p between 0
and 1, the (100p)th percentile of the distribution is the number πp such that
πp
(integral)
−∞. f(x) dx= p.
Thus, π0.50 is the median and π0.23 is the 23rd percentile.
(a) Find the values of π0.005, π0.025, π0.05, π0.95, π0.975, π0.995 for a N(0,1) distribution.
(b) If Z∼N(0,1), find the values of P(π0.005 <Z<π0.995), P(π0.025 <Z<π0.975), and P(π0.05 <Z<π0.95).
m
(integral)
−∞. f(x) dx= 0.5.
That is, if X is a random variable with this distribution, then P(X≤m) = 0.5.
The median is also known as the 50th percentile. We can also consider other
percentiles. For example, if we want the 23rd percentile, then we try to find a
number q such that
q
(integral)
−∞ f(x) dx= 0.23.(1)
The general notation used by the textbook is that for any number p between 0
and 1, the (100p)th percentile of the distribution is the number πp such that
πp
(integral)
−∞. f(x) dx= p.
Thus, π0.50 is the median and π0.23 is the 23rd percentile.
(a) Find the values of π0.005, π0.025, π0.05, π0.95, π0.975, π0.995 for a N(0,1) distribution.
(b) If Z∼N(0,1), find the values of P(π0.005 <Z<π0.995), P(π0.025 <Z<π0.975), and P(π0.05 <Z<π0.95).
