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lubasha [3.4K]
3 years ago
10

A food processor packages orange juice in small jars. The weights of the filled jars are approximately normally distributed with

a mean of 10.5 ounces and a standard deviation of 0.3 ounce. Find the proportion of all jars packaged by this process that have weights that fall above 10.983 ounces.
Mathematics
1 answer:
Ratling [72]3 years ago
8 0

Answer:

P(X>10.983)=P(\frac{X-\mu}{\sigma}>\frac{10.983-\mu}{\sigma})=P(Z>\frac{10.983-10.5}{0.3})=P(z>1.61)

And we can find this probability using the complement rule and with excel or the normal standard table:

P(z>1.61)=1-P(z

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the weights of a population, and for this case we know the distribution for X is given by:

X \sim N(10.5,0.3)  

Where \mu=10.5 and \sigma=0.3

We are interested on this probability

P(X>10.983)

And the best way to solve this problem is using the normal standard distribution and the z score given by:

z=\frac{x-\mu}{\sigma}

If we apply this formula to our probability we got this:

P(X>10.983)=P(\frac{X-\mu}{\sigma}>\frac{10.983-\mu}{\sigma})=P(Z>\frac{10.983-10.5}{0.3})=P(z>1.61)

And we can find this probability using the complement rule and with excel or the normal standard table:

P(z>1.61)=1-P(z

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