Question

a chocolate truffle machine makes truffles with normally distributed weights with mean 22.2 and standard deviation...

a chocolate truffle machine makes truffles with normally distributed weights with mean 22.2 and standard deviation 3.0.

a. if you take a random truffle produced by the machine, find the probability it weighs more than 24

b. the company makes small gift boxes that contain 4 truffles. if you are given one of these boxes, what is the probability that the total weight of the truffles is more than 96?

c. suppose some adjustments are made to the machine so that standard deviation is still 3.0, but the mean weight is no longer known. To estimate the new mean weight, you take an SRS of 100 truffles and find the mean of the weights in the sample is 21. Find a 99% confidence interval for the mean weight of all truffles produced by the updated machine.

d. how can the company improve their estimate of the mean in part c?

e. the machine breaks and is replaced by a totally different machine of which you have no knowledge about mean or standard deviation of the truffles. You take an SRS of 2500 truffles, find a mean weight of 23.1 and standard deviation of 2.5. estimate the mean weight of all truffles produced by the new machine using a 99% confidence interval.

f. this new machine occasionally malfunctions and produces a truffle that cannot be sold. To estimate how often this can happen, you take a SRS of 300 truffles and find that 3 cannot be sold. estimate how often truffles can not be sold using a 95% confidence interval.

Homework Answers

Answer #1

olution-A:

P(Xbar>24)

z=xbar-mu/sigma/qrt(n)

z=24-22.2/((3/sqrt(1)

z=0.6

P(Z>0.6)

=1-P(Z<0.6)

=1-0.7257

=0.2743

0.2743

Solution-b:

s=sigma/sqrt(n)=3/sqrt(4)=3/2=1.5

xbar=22.2

P(xbar>96)

P(Z>96-22.2/1.5)

P(Z> 49.2)

=1-P(Z<49.2)

=1-1

=0

0.0000

Solution-c:

since popualtion standard deviation is given use z critical

z critical for 99%=2.576

sigma=3

xbar=21

n=100

99% confidence interval for mean is

xbar-z*sigma/sqrt(n),xbar+z*sigma/sqrt(n)

21-2.576*3/sqrt(100),21+2.576*3/sqrt(100)

20.2272, 21.7728

99% lower limit mean=20.2272

99% upper limit mean=21.7728

d. how can the company improve their estimate of the mean in part c

cmpany improve their estimate of the mean in part c by increasing sample size

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