SOLUTION: Let the random variable X follow a normal distribution with mean μ = 50 and variance σ2 = 25. a) Find the probability that X is greater than 58. b) Find the probabilit

Algebra ->  Probability-and-statistics -> SOLUTION: Let the random variable X follow a normal distribution with mean μ = 50 and variance σ2 = 25. a) Find the probability that X is greater than 58. b) Find the probabilit      Log On


   



Question 380422: Let the random variable X follow a normal distribution with mean μ = 50 and
variance σ2 = 25.
a) Find the probability that X is greater than 58.
b) Find the probability that X is greater than 46 and less than 60.
c) If you draw a sample of size n=20 from the X population described above,
what is the probability that the sample variance, s2, is larger than 30?
d) Imagine a symmetric interval about the mean (μąc) of the distribution
described above. Find the value for c such that the probability is
approximately 0.1 that X falls in this interval.

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Let the random variable X follow a normal distribution with mean μ = 50 and
variance σ2 = 25.
Note: std = 5
=======================
a) Find the probability that X is greater than 58.
z(58) = (58-50)/5 = 8/5 = 1.4
P(x> 58) = P(z> 1.4) =
--------------------------------
b) Find the probability that X is greater than 46 and less than 60.
Do the same as above for 46 and 60.
------------------------------------------------
c) If you draw a sample of size n=20 from the X population described above,
what is the probability that the sample variance, s2, is larger than 30?
std of the sample means = 5/sqrt(20)
---
t(30) = (30-50)/[5/sqrt(20)] = -17.89
----
P(x-bar > 30) = P(t > -17.89 when df = 19) = tcdf(-17.89,100,19) = 1.00
----------------------------
d) Imagine a symmetric interval about the mean (μąc) of the distribution
described above. Find the value for c such that the probability is
approximately 0.1 that X falls in this interval.
----
Draw the picture using a normal curve.
You have an interval with 0.05 to the left and to the right of the mean.
That leaves a left-tail of 0.45 to the left.
z = +/-invNorm(0.45) = +-0.1257
---------------
Left boundary:
c = -0.1257*5+50 = 49.37
---
Right boundary:
c = +0.1257*5+50 = 50.63
=============================
Cheers,
Stan H.