Question

. A soft-drink machine is regulated so that it discharges an average of 200 milliliters per cup. If the amount of drink is normally distributed with a standard deviation equal to 15 milliliters, (a) what fraction of the cups will contain more than 224 milliliters? (b) what is the probability that a cup contains between 191 and 209 milliliters? (c) below what value do we get the smallest 25% of the drinks

Answer 1

I had this question in my book

Answer 2

Answer:

Sample mean (known as xbar)=200  

sd=15  

x=fraction of cups  

P(x > 224)  

compute z=(x-xbar)/sd  

z=(224-200)/15=24/15=8/5=1.6  

a)P(x > 224)=p(z > 1.6)=0.0548=548/10000  

You'll need a normal probability table. If you don't have one, the source gives the normal probability. Note that we want the probability of z exceeding 1.6. Draw a picture and shade the area in each case, so it'll be easy to know what area we want.  

b)between 191 and 209  

convert them both to z scores  

(191-200)/15=-9/15=-3/5=-0.6  

(209-200)/15=9/15=3/5=0.6  

P( 191 < x < 209)=P( -0.6 < z < 0.6)=0.2257(2)=0.4514  

c)P(x > 230)=p(z >(230-200)/15)=p(z > 2)=0.0228  

multiply this by 1000=22.8 or 23 cups will overflow out of 1000.  

d)Find value such that p( z < value)=0.25  

This value is -0.67. This is a z value.  

-0.67=(x-200)/15  

solve for x  

x=200-0.67(15)=109.95 below which we get the smallest 25 % of the drinks

Step-by-step explanation: