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goldenfox [79]
3 years ago
14

The weight of a package of mints is believed to be normally distributed with mean 21.37 grams and standard deviation 0.4, i.e. X

\sim N(21.37,0.4). What is the chance that a sample of 4 packages of mints has an average weight, \overline X, between 21 and 22 grams
Mathematics
1 answer:
nasty-shy [4]3 years ago
6 0

Answer:

P(21

And using a calculator, excel or the normal standard table we have that:

P(-1.85

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 weigths of a population, and for this case we know the distribution for X is given by:

X \sim N(21.37,0.4)  

Where \mu=21.37 and \sigma=0.4

Since the distribution for X is normal then we know that the distribution for the sample mean \bar X is given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

We can find the probability required with the following z score formula:

z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}


And replacing we got:

P(21

And using a calculator, excel or the normal standard table we have that:

P(-1.85

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