Question 999952: Let X be a normal random variable with mean 208 units and standard deviation 3 units. Answer the following questions, rounding your answers to two decimal places.
(a) What is the probability that X will be less than 201 units, P(X<201)?
(b) What is the probability that X will be within 7 units of the mean?
(c) The probability is 0.1 that X will be more than how many units?
Thank you!
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Let X be a normal random variable with mean 208 units and standard deviation 3 units. Answer the following questions, rounding your answers to two decimal places.
(a) What is the probability that X will be less than 201 units, P(X<201)?
z(201) = (201-208)/3 = -7/3
P(x < 201) = P(z < -7/3) = normalcdf(-100,-7/3) = 0.0098
-----------------------------------------
(b) What is the probability that X will be within 7 units of the mean?
P(201< x <215) = normalcdf(201,215,208,3) = 0.9804
-----------------------------------------
(c) The probability is 0.1 that X will be more than how many units?
Find the z-value with a left tail of 0.90
invNorm(0.90) = 1.2816
-----
Find the corresponding x-value::
x = 1.2816*3+208
x = 211.84
---------------------
Cheers,
Stan H.
-------------
|
|
|
