Question 310959
60% of the students in the computer course passed the test. if 25 students were randomly chosen at random.
----
Binomial Problem with n = 25 ; p = 0.6
---- 
1) what is the probabiliy that more than 18 students passed the test
P(19 <= x <= 25) = 1 - P(0<= x <=18) = 1-binomcdf(25,0.6,18) = 0.0736
----------------------------------------------------------- 
2)what is the probability between 10 and 17 passed the test
binomcdf(25,0.6,17) - binomcdf(25,0.6,9) = 0.8333
----------------------------------------------------------- 
3) calculate the expected mean = np = 25*0.6 = 15
-------------------- 
4)calculate the standard deviation = sqrt(npq) = sqrt(15*0.4) = 2.45
==================================================================== 
A soft drink bottling company wants to fill bottles labeled 360 ml such that bottles that contain less than 355.5 ml will be declared as non-quality standard and sold st substantial discount. The fills are normally distributed with a mean of 360 ml and standard deviation of 3.50ml. If 20,000 bottles are filled, approximatly HOW MANY will be declared as non-quality standard
----
Draw the picture of a normal curve
Put 360 in the middle and note that std = 3.50
Find the z value of 355.5: 
z(355.5) = (355.5-360)/3.5 = -1.2857
--
Then P(z < -1.2857) = 0.0993
------------------------------------
If 20,000 bottles are filled, approximatly HOW MANY will be declared as non-quality standard
Ans: 0.0993*20000 = 1985
==============================
Cheers,
Stan H.
==============================