Mathematics, 19.02.2020 21:24 janiyahmcgolley

The distribution of weights of United States pennies is approximately normal with a mean of 2.5 grams and a standard deviation of 0.03 grams.

a. What is the probability that a randomly chosen penny weights less than 2.4 grams?

b. Describe the sampling distribution of the mean weight of 10 randomly chosen pennies.

c. What is the probability that the mean weight of 10 pennies is less than 2.4 grams?

d. Sketch the two distributions (population and sampling) on the same scale.

e. Could you estimate the probabilities from (a) and (c) if the weights of the pennies had a skewed distribution?

Answers: 3

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Answer from: lashaaungas

a) P(X

b) Since the distribution for X is normal then the distribution for the sample mean $\bar X$ is also normal and given by:

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

$\mu_{\bar X}=2.5$

$\sigma_{\bar X}=\frac{0.03}{\sqrt{10}}=0.00949$

c) $P(\bar X$

We can use the z score formula given by $z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And using the z score formula we got:

$P(\bar X$

d) Figure attached.

As we can see we have less deviation for the sample mean and for this reason we have more values near to the mean.

e) Since we just have the mean and the deviation we can't estimate the probability if we don't know the distribution.

The z score can't be applied if we have a normal distribution.

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".

Part a

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:

$X \sim N(2.5,0.03)$

Where $\mu=2.5$ and $\sigma=0.03$

And we want this probability:

P(X

We can use the z score formula given by $z = \frac{x -\mu}{\sigma}$

And using the z score formula we got:

P(X

Part b

Since the distribution for X is normal then the distribution for the sample mean $\bar X$ is also normal and given by:

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

$\mu_{\bar X}=2.5$

$\sigma_{\bar X}=\frac{0.03}{\sqrt{10}}=0.00949$

Part c

$P(\bar X$

We can use the z score formula given by $z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And using the z score formula we got:

$P(\bar X$

Part d

Figure attached.

As we can see we have less deviation for the sample mean and for this reason we have more values near to the mean.

Part e

Since we just have the mean and the deviation we can't estimate the probability if we don't know the distribution.

The z score can't be applied if we have a normal distribution.

The distribution of weights of United States pennies is approximately normal with a mean of 2.5 gram

Answer from: Quest

29 is the correct answer

step-by-step explanation:

Answer from: Quest

25 questions , because they start with 100.

350-100=250 .

. 250/10=25

step-by-step explanation:

first take off 100, then what x 10 = 250? that is your answer. (25)

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The distribution of weights of United States pennies is approximately normal with a mean of 2.5 gram...