Question

The weights of the fish in a certain lake are normally distributed with a mean of 12 lb and a standard deviation of 6. If 4 fish are randomly​ selected, what is the probability that the mean weight will be between 9.6 and 15.6 ​lb? Round your answer to four decimal places.

Answer 1

Answer:

The probability that the mean weight will be between 9.6 and 15.6 ​lb is 0.6731

Step-by-step explanation:

Here we have

P(9.6≤$\bar{x}$≤15.6)

Therefore, we have

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

Where:

$\bar{x}$ = Sample mean

μ = Population mean = 12

σ = Population standard deviation = 6

n = Sample size = 4

When $\bar{x}$ = 9.6 we have

$z=\frac{9.6-12 }{\frac{6 }{\sqrt{4}}}$ = -0.8

From Z table, that is 0.21186

When $\bar{x}$ = 15.6 we have 1.2

$z=\frac{15.6-12 }{\frac{6 }{\sqrt{4}}}$ = 1.2

From Z table, that is 0.88493

0.88493 - 0.21186 = 0.67307

P(9.6≤$\bar{x}$≤15.6) = 0.67307 which is 0.6731 to four decimal places.

Answer 2

Answer:

$P(9.6 < \bar X$

And using the normal standard distribution or excel we got:

$P(9.6 < \bar X$

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:

$X \sim N(12,6)$  

Where $\mu=12$ and $\sigma=6$

Since the dsitribution 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 want to find this probability:

$P(9.6 < \bar X$

And we can use the z score formula given by;

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

And if we find the z score for the limits given we got:

$z = \frac{9.6-12}{\frac{6}{\sqrt{4}}}= -0.8$

$z = \frac{15.6-12}{\frac{6}{\sqrt{4}}}= 1.2$

So we can calculate this probability like this:

$P(9.6 < \bar X$

And using the normal standard distribution or excel we got:

$P(9.6 < \bar X$