Question

The amount of water in a bottle is approximately normally distributed with a mean of 2.55 liters with a standard deviation of 0.035 liter. Complete parts (a) through (d) below. a. What is the probability that an individual bottle contains less than 2.52 liters? Round to three decimal places as needed.) b. If a sample of 4 bottles is selected, what is the probability that the sample mean amount contained is less than 2.52 liters? (Round to three decimal places as noeded) c. If a sample of 25 bottles is selected, what is the probability that the sample mean amount contained is less than 2.52 liters? (Round to three decimal places as needed.) d. Explain the difference in the results of (a) and (c)

Answer 1

Answer:

a) $P(X<2.52)=P(Z<\frac{2.52-2.55}{0.035})=P(Z<-0.857)=0.196$

b) $P(\bar X <2.52) = P(Z<-1.714)=0.043$

c) $P(\bar X <2.52) = P(Z<-4.286)=0.000$

d) For part a we are just finding the probability that an individual bottle would have a value of 2.52 liters or less. So we can't compare the result of part a with the results for parts b and c.

If we see part b and c are similar but the difference it's on the sample size for part b we just have a sample size 4 and for part c we have a sample size of 25. The differences are because we have a higher standard error for part b compared to part c.

Step-by-step explanation:

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".  The letter $\phi(b)$ is used to denote the cumulative area for a b quantile on the normal standard distribution, or in other words: $\phi(b)=P(z<b)$

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

$X \sim N(2.55,0.035)$

a. What is the probability that an individual bottle contains less than 2.52 liters?

We are interested on this probability

$P(X<2.52)$

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:

$P(X<2.52)=P(\frac{X-\mu}{\sigma}<\frac{2.52-\mu}{\sigma})$

And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.  

$P(X<2.52)=P(Z<\frac{2.52-2.55}{0.035})=P(Z<-0.857)=0.196$

b. If a sample of 4 bottles is selected, what is the probability that the sample mean amount contained is less than 2.52 liters? (Round to three decimal places as noeded)

And let $\bar X$ represent the sample mean, the distribution for the sample mean is given by:

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

On this case  $\bar X \sim N(2.55,\frac{0.035}{\sqrt{4}})$

The z score on this case is given by this formula:

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

And if we replace the values that we have we got:

$z=\frac{2.52-2.55}{\frac{0.035}{\sqrt{4}}}=-1.714$

For this case we can use a table or excel to find the probability required:

$P(\bar X <2.52) = P(Z<-1.714)=0.043$

c. If a sample of 25 bottles is selected, what is the probability that the sample mean amount contained is less than 2.52 liters? (Round to three decimal places as needed.)

The z score on this case is given by this formula:

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

And if we replace the values that we have we got:

$z=\frac{2.52-2.55}{\frac{0.035}{\sqrt{25}}}=-4.286$

For this case we can use a table or excel to find the probability required:

$P(\bar X <2.52) = P(Z<-4.286)=0.0000091$

d. Explain the difference in the results of (a) and (c)

For part a we are just finding the probability that an individual bottle would have a value of 2.52 liters or less. So we can't compare the result of part a with the results for parts b and c.

If we see part b and c are similar but the difference it's on the sample size for part b we just have a sample size 4 and for part c we have a sample size of 25. The differences are because we have a higher standard error for part b compared to part c.  

Answer 2

The expected value are given for the population, and as the sample

increase, the sample mean approaches the population mean.

Responses:

a. 0.195

b. 0.044

c. 0

d. The sample mean approaches the population mean as the sample size  gets larger

How can the probability from the z-score of a value?

Given:

The mean, μ = 2.55 liters

The standard deviation, σ = 0.035 liter

The z-score is given by the formula;

$Z= \mathbf{\dfrac{x-\mu }{\sigma }}$

a. The probability that an individual bottle contains less than 2.52 liters

is therefore;

$P(x < 2.52) = P\left(Z<\dfrac{2.52-2.55 }{0.035 }\right) \approx P\left(Z< -0.857\right) \approx \mathbf{0.195}$

• The probability that an individual bottle contains less than 2.52 liters is approximately 0.195

b. The z-score for a given sample mean, $\mathbf{\overline x}$ is given by the formula;

$Z= \mathbf{\dfrac{\bar{x}-\mu }{\frac{\sigma }{\sqrt{n}}}}$

Which gives;

$Z= \mathbf{\dfrac{ 2.52 - 2.55 }{\frac{0.035 }{\sqrt{4}}}} \approx -1.714$

$P\left(Z< -1.714\right) \approx 0.044$

• The probability that the sample mean of 4 bottles is less than 2.52 liters is approximately 0.044

c. Where the number of bottles selected is 25 bottles, we have;

$z= \mathbf{\dfrac{ 2.52 - 2.55 }{\frac{0.035 }{\sqrt{25}}}} \approx -4.286$

$\mathbf{P\left(Z< -4.286\right)} \approx 0$

• The probability that the mean amount contained in 25 bottles is less than 2.52 is approximately 0

d. The results indicate that as the number of bottles selected increase,

the mean amount water contained approaches 2.55 liters, which is as

stated and given by the Central Limit Theorem.

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