Respuesta :
Answer:
[tex] P(9.6 < \bar X <15.6) =P(-0.8<z<1.2)= P(Z<1.2)- P(Z<-0.8)[/tex]
And using the normal standard distribution or excel we got:
[tex] 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 [/tex]
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:
[tex]X \sim N(12,6)[/tex]
Where [tex]\mu=12[/tex] and [tex]\sigma=6[/tex]
Since the dsitribution for x is normal then we know that the distribution for the sample mean [tex]\bar X[/tex] is given by:
[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]
We want to find this probability:
[tex] P(9.6 < \bar X <15.6)[/tex]
And we can use the z score formula given by;
[tex] z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
And if we find the z score for the limits given we got:
[tex] z = \frac{9.6-12}{\frac{6}{\sqrt{4}}}= -0.8[/tex]
[tex] z = \frac{15.6-12}{\frac{6}{\sqrt{4}}}= 1.2[/tex]
So we can calculate this probability like this:
[tex] P(9.6 < \bar X <15.6) =P(-0.8<z<1.2)= P(Z<1.2)- P(Z<-0.8)[/tex]
And using the normal standard distribution or excel we got:
[tex] 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 [/tex]
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≤[tex]\bar{x}[/tex]≤15.6)
Therefore, we have
[tex]z=\frac{\bar{x}-\mu }{\frac{\sigma }{\sqrt{n}}}[/tex]
Where:
[tex]\bar{x}[/tex] = Sample mean
μ = Population mean = 12
σ = Population standard deviation = 6
n = Sample size = 4
When [tex]\bar{x}[/tex] = 9.6 we have
[tex]z=\frac{9.6-12 }{\frac{6 }{\sqrt{4}}}[/tex] = -0.8
From Z table, that is 0.21186
When [tex]\bar{x}[/tex] = 15.6 we have 1.2
[tex]z=\frac{15.6-12 }{\frac{6 }{\sqrt{4}}}[/tex] = 1.2
From Z table, that is 0.88493
0.88493 - 0.21186 = 0.67307
P(9.6≤[tex]\bar{x}[/tex]≤15.6) = 0.67307 which is 0.6731 to four decimal places.
