Answer to Question #224231 in Statistics and Probability for Arham

Question #224231
A company has designed a coffee vending machine which fills 210 ml coffee in each cup, on average. You are supposed to perform find some statistics as a part of quality control department. If the amount of drink dispensed regularly from this machine is normally distributed with a standard deviation equal to 13.5 ml,
a) What percentage of the cups will contain more than 231 ml?

b) What would be the chance that a cup will contain between 180 and 220 ml?

c) Calculate the number of cups that will have a probability to overflow if 235 ml cups are used for 900 drinks sold at any random day?

d) If only less than 15% cups are overflowed, what would the cutoff filling volume be?
1
Expert's answer
2021-08-09T13:12:09-0400

Let X=X= the amount of coffee in a cup: XN(μ,σ2).X\sim N(\mu, \sigma^2).

Given μ=210 ml,σ=13.5 ml.\mu=210\ ml, \sigma=13.5\ ml.

a)


P(X>231)=1P(Z23121013.5)P(X>231)=1-P(Z\leq\dfrac{231-210}{13.5})

=1P(Z149)0.0599=1-P(Z\leq \dfrac{14}{9})\approx0.0599

6% of the cups will contain more than 231 ml.


b)


P(180<X<220)=P(X<220)P(X180)P(180<X<220)=P(X<220)-P(X\leq180)

=P(Z<22021013.5)P(Z18021013.5)=P(Z<\dfrac{220-210}{13.5})-P(Z\leq\dfrac{180-210}{13.5})




0.77057470.01313410.757441\approx0.7705747-0.0131341\approx0.757441

0.7574410.757441


c)

P(X>235)=1P(Z23521013.5)P(X>235)=1-P(Z\leq\dfrac{235-210}{13.5})

=1P(Z5027)0.032024=1-P(Z\leq\dfrac{50}{27})\approx0.032024

0.032024(900)=290.032024(900)=29

2929 cups


d)


P(X>x)=0.15P(X>x)=0.15P(X>x)=1P(Zx21013.5)=0.15P(X>x)=1-P(Z\leq\dfrac{x-210}{13.5})=0.15

P(Zx21013.5)=0.85P(Z\leq\dfrac{x-210}{13.5})=0.85

x21013.51.036433\dfrac{x-210}{13.5}\approx1.036433

x=224x=224

224224 ml.




Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment