Respuesta :
Answer:
[tex]P(X<6)=P(\frac{X-\mu}{\sigma}<\frac{6-\mu}{\sigma})=P(Z<\frac{6-6}{0.45})=P(z<0)[/tex]
And we can find this probabiity using the normal standard table or excel and we got:
[tex]P(z<0)=0.5[/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(6,0.45)[/tex]
Where [tex]\mu=6[/tex] and [tex]\sigma=0.45[/tex]
We are interested on this probability
[tex]P(X<6)[/tex]
And the best way to solve this problem is using the normal standard distribution and the z score given by:
[tex]z=\frac{x-\mu}{\sigma}[/tex]
If we apply this formula to our probability we got this:
[tex]P(X<6)=P(\frac{X-\mu}{\sigma}<\frac{6-\mu}{\sigma})=P(Z<\frac{6-6}{0.45})=P(z<0)[/tex]
And we can find this probabiity using the normal standard table or excel and we got:
[tex]P(z<0)=0.5[/tex]
50% of rats have a weight of less than 6 lbs.
Z score is used to determine by how many standard deviations the raw score is above or below the mean. The z score is given by:
[tex]z=\frac{x-\mu}{\sigma} \\\\where\ x=raw\ score,\mu=mean,\sigma=standard\ deviation[/tex]
Given that μ = 6 lbs, σ = 0.45 lbs.
For x < 6:
[tex]z=\frac{6-6}{0.45}=0[/tex]
From the normal distribution table, P(x < 6) = P(z < 0) = 0.5 = 50%
50% of rats have a weight of less than 6 lbs.
Find out more at: https://brainly.com/question/15016913
