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Sati [7]
4 years ago
6

A filling machine puts an average of four ounces of coffee in jars, with a standard deviation of 0.25 ounces. forty jars filled

by this machine are selected at random. what is the probability that the mean amount per jar filled in the sampled jars is less than 3.9 ounces
Mathematics
1 answer:
Lunna [17]4 years ago
4 0
Let's attack this problem using the z-score concept.  The sample std. dev. here is (0.25 oz)/sqrt(40), or 0.040.  Thus, the z score representing 3.9 oz. is
        3.9 - 4.0
z = -------------- = -2.5
          0.040 

In one way or another we must find the area under the std. normal curve that lies to the left of z = -2.5.  Use a table of z-scores or a calculator with built-in statistics functions.  According to my TI-83 Plus calculator, that area is

0.006.  One way of interpreting this that with so small a standard deviation, most volumes of coffee put into the jars are very close to the mean, 4 oz.
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Subtract the following:

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B) 49 rupees 79 paise from 123 rupees 68 paise.

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Hint: Use decimal concept.

We know that, 1 rupee = 100 paise. We can reframe these questions as follows:

18 rupees 9 paise from 75 rupees 80 paise

18 rupees 9 paise can be represented as 18.09 rupees and 75 rupees 80 paise can be represented as 75.80 rupees. Now, on subtracting we’ll get,

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49 rupees 79 paise can be represented as 49.79 rupees and 123 rupees 68 paise can be represented as 123.68 rupees. Now, on subtracting we’ll get,

123.68 −49.79−−−−−− 73.89

Which means 73 rupees 89 paise.

Note: We can also perform the subtraction by making all units the same that are paise and then subtract.

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