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Chocolate bars produced by a certain machine are labeled with 8.0 oz. The distribution of the acutal weights of these chocolates bars is Normal with a mean of 8.1 oz and a standard deviation of 0.1oz. A chocolate bar is considered underweight if it weighs less then 8.0 oz.

(A) What proportion of chocolate bars wieghs less then 8.0 oz?
(B) What propertion of chocolate bars weighs between 8.2 and 8.3 oz?
(C) How should the chocolate bar wrappers be labeled so that only 1% of such bars are underwight?

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Answer:

[tex]a.\ P(C<8.0)=0.1587\\\\b. \ P(8.2<C<8.3)=0.1359\\\\c. 7.87\ oz[/tex]

Step-by-step explanation:

a. Let C be the normally distributed random variable.

-Given the chocolates is normally distributed with mean =8.1 oz and standard deviation =0.1 oz.

#The proportion of those less than 8.0 oz can be calculated as:

[tex]P(C<8.0)=P(Z<\frac{\bar X-\mu}{\sigma})\\\\\\=P(Z<\frac{8.0-8.1}{0.1})\\\\\\=P(Z<-1)=0.1587[/tex]

Hence, the proportion of bars weighing below 8 ounces is 0.1587 or 15.87%

b. We use the z-test to determine the probability or proportion of bars weighing between 8.2 and 8.3 ounces:

[tex]P(8.2<C<8.3)=P(\frac{\bar X-\mu}{\sigma}<Z<\frac{\bar X-\mu}{\sigma})\\\\\\=P(\frac{8.2-8.1}{0.1}<Z<\frac{8.3-8.1}{0.1})\\\\\\=P(1<Z<2)\\\\\\=0.9772-0.8413\\\\=0.1359[/tex]

Hence, the proportion of bars weighing between 8.2 and 8.3 ounces is 0.1359 or 13.59%

c. To label the wrappers such that only 1% are underweight.

#Find B such that

P(C<B)=0.01

#Now find z-value such that:

[tex]P(Z<B)=0.01[/tex]

Using the z-tables, z=-2.33

Therefore:

[tex]C=z\sigma +\mu\\\\=-2.333\times 0.1+8.1\\\\=7.8667\approx 7.87 \ oz[/tex]

Hence, the wrappers should be labelled as 7.87 ounces

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