Question

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?

Answer 1

Answer:

a) $P(X$

b) $\bar X \sim N(2.5, \frac{0.03}{\sqrt{10}}=0.00948)$

c) $P(\bar X$

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

$P(Z$

d) Figure attached

e) If we don't know the distribution then we can't ensure that the sample mean would be distributed like this:

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

And we can't estimate the probabilities on a easy way.

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 $\sigma=0.03$

We are interested on this probability

$P(X$

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:

$P(X$

And we can find this probability on this way:

$P(-0.50$

Part b

Since the distribution for X is normal then the distribution for the sample mean is:

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

$\bar X \sim N(2.5, \frac{0.03}{\sqrt{10}}=0.00948)$

Part c

$P(\bar X$

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

$P(Z$

Part d

See the figure attached the deviation for the sample mean is lower for this reason we have the pattern in the graph attached.

Part e

If we don't know the distribution then we can't ensure that the sample mean would be distributed like this:

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

And we can't estimate the probabilities on a easy way.