Question 673195
the chances of any detergent having at least 2.4grams of this chemical are 2.28% while the chances of having less than 1.4grams are 30.85%
---------
The z-value with a right tail of 2.28% is z = invNorm(0.9772) = 1.9991
The z-value with a left tail of 30.85% is z = invNorm(0.3085) = 0.5001
Equations based on x = z*s + u
2.4 = 1.9991*s + u
1.4 = 0.5001*s + u
-------
Subtract to solve for "s":
1 = 1.499s
s = 0.667
----
Solve for "u":
2.4 = 1.9991*0.667 + u
u = 1.0666


a) determine the probability of a detergent having
i)more than 1.2grams but not exceeding 2.6grams of the chemical
z(1.2) = (1.2-1.0666)/0.667 = 0.1999
z(2.6) = (2.6-1.0666)/0.667 = 2.2990
-----
P(1.2 < x < 2.6) = P(0.1999 < z < 2.2990) = 0.4100
---------------------------------------------------------

ii)
either less than 0.4grams
z(0.4) = (0.4-1.0666)/0.667 = -0.999
P(x < 0.4) = P(z < -0.999) = normalcdf(-100,-0.999) = 0.1589
---------------- 

exactly 0.8grams
P(x = 0.8) = 0
Why?:: The probability of ANY particular value in a continuous
distribution is zero.
-------------------------

more than 2.8grams
z(2.8) = (2.8-1.0666)/0.667 = 2.5987
P(x > 2.8) = P(z > 2.5987) = normalcdf(2.5987,100) = 0.0047
===============================================================
cheers,
Stan H.
=======================