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

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

Explanation:

Here we have

P(9.6≤
`\bar{x}`≤15.6)

Therefore, we have

`z=\frac{\bar{x}-\mu }{(\sigma )/(√(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=(9.6-12 )/((6 )/(√(4)))` = -0.8

From Z table, that is 0.21186

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

`z=(15.6-12 )/((6 )/(√(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

`P(9.6 < \bar X <15.6) =P(-0.8<z<1.2)= P(Z<1.2)- P(Z<-0.8)`

And using the normal standard distribution or excel we got:

`P(9.6 < \bar X <15.6) =P(-0.8<z<1.2)= P(Z<1.2)- P(Z<-0.8)=0.8849- 0.2119=0.6731`

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, (\sigma)/(√(n)))`

We want to find this probability:

`P(9.6 < \bar X <15.6)`

And we can use the z score formula given by;

`z = (\bar X -\mu)/((\sigma)/(√(n)))`

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

`z = (9.6-12)/((6)/(√(4)))= -0.8`

`z = (15.6-12)/((6)/(√(4)))= 1.2`

So we can calculate this probability like this:

`P(9.6 < \bar X <15.6) =P(-0.8<z<1.2)= P(Z<1.2)- P(Z<-0.8)`

And using the normal standard distribution or excel we got:

`P(9.6 < \bar X <15.6) =P(-0.8<z<1.2)= P(Z<1.2)- P(Z<-0.8)=0.8849- 0.2119=0.6731`