Respuesta :
Answer:
[tex]z=-1.995<\frac{a-10}{5}[/tex]
And if we solve for a we got
[tex]a=10 -1.995*5=0.025[/tex]
Step-by-step explanation:
For this case we have the initial case we have a normal distribution given :
[tex]X \sim N(0,5)[/tex]
Where [tex]\mu=0[/tex] and [tex]\sigma=5[/tex]
And from the info of the problem we know that:
[tex] P(X<-10) =0.023[/tex]
We can verify this using the z score formula given by:
[tex] z= \frac{x\mu}{\sigma}[/tex]
And if we replace we got:
[tex] P(X<-10) = P(Z<\frac{-10-0}{5}) = P(Z<-2) = 0.02275 \approx 0.023[/tex]
Now we have another situation for a new random variable X with a new distribution given by:
[tex]X \sim N(10,5)[/tex]
Where [tex]\mu=10[/tex] and [tex]\sigma=5[/tex]
For this part we want to find a value a, such that we satisfy this condition:
[tex]P(X>a)=0.977[/tex] (a)
[tex]P(X<a)=0.023[/tex] (b)
Both conditions are equivalent on this case. We can use the z score again in order to find the value a.
As we can see on the figure attached the z value that satisfy the condition with 0.023 of the area on the left and 0.977 of the area on the right it's z=-1.995. On this case P(Z<-1.995)=0.023 and P(z>-1.995)=0.977
If we use condition (b) from previous we have this:
[tex]P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.023[/tex]
[tex]P(z<\frac{a-\mu}{\sigma})=0.023[/tex]
But we know which value of z satisfy the previous equation so then we can do this:
[tex]z=-1.995<\frac{a-10}{5}[/tex]
And if we solve for a we got
[tex]a=10 -1.995*5=0.025[/tex]
Using the normal distribution and the concept of z-score, it is found that 0.023 of the curve will fall below X = 0.02.
In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
- It measures how many standard deviations the measure is from the mean.
- After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X, and also the area under the curve to the left Z.
In this problem:
- The mean is 10 and the standard deviation is 5, thus [tex]\mu = 10, \sigma = 5[/tex].
- Area to the left under the normal curve of 0.023, thus Z with a p-value of 0.023, that is, [tex]Z = -1.996[/tex].
Then
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]-1.996 = \frac{X - 10}{5}[/tex]
[tex]X - 10 = 5(-1.996)[/tex]
[tex]X = 0.02[/tex]
The score is 0.02.
A similar problem is given at https://brainly.com/question/25154104
