Question

The weight, in grams, of coffee beans in a bag is normally distributed with a mean and a standard deviation of 7.8. Given that 10% of bags contain less than 200g, find:

a.) the value of the mean

b.) the percentage of bags that contain more than 215g.

I do not understand the answer that has already been posted for this same question. Please explain better.

Answer 1

Define X= weight in grams.

Given that X~N(m,SD=7.8), m is unknown.

a) Since 10% of bags contain less than 200g, we have P(X<200)=.10 or P(Z<(200-m)/7.8)=.10, where Z=(X-m)/7.8~N(0,1).

Using normal tables, we find that P(Z<-1.281552)=.10 and hence

P(Z<(200-m)/7.8)=.10=P(Z<-1.281552) so that (200-m)/7.8=-1.281552 or m=200+7.8*1.281552=209.9961

Henne mean=209.9961g.
b) Percentage of bags that contain more than 215g= 100P(X>215)%

Now P(X>215)=P(Z>(215-m)/7.8)=P(Z>0.6415256)=1-P(Z<=0.6415256)=1-0.7394094=0.2605906 (using Normal tables)

Thus the required % is 26.0591%.

For any query in above, comment.