Respuesta :
Answer:
P(x >92) = 0.1056
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score 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. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
[tex]\mu = 82, \sigma 8[/tex]
P(x >92)
This is 1 subtracted by the pvalue of Z when X = 92. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{92 - 82}{8}[/tex]
[tex]Z = 1.25[/tex]
[tex]Z = 1.25[/tex] has a pvalue of 0.8944
1 - 0.8944 = 0.1056
P(x >92) = 0.1056
Answer:
[tex]P(X>92)=P(\frac{X-\mu}{\sigma}>\frac{92-\mu}{\sigma})=P(Z>\frac{92-82}{8})=P(z>1.25)[/tex]
And we can find this probability with the complement rule and using the normal standard table or excel:
[tex]P(z>1.25)=1-P(z<1.25)=1-0.894=0.106 [/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 final exam grades of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(82,8)[/tex]
Where [tex]\mu=82[/tex] and [tex]\sigma=8[/tex]
We are interested on this probability
[tex]P(X>92)[/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>92)=P(\frac{X-\mu}{\sigma}>\frac{92-\mu}{\sigma})=P(Z>\frac{92-82}{8})=P(z>1.25)[/tex]
And we can find this probability with the complement rule and using the normal standard table or excel:
[tex]P(z>1.25)=1-P(z<1.25)=1-0.894=0.106 [/tex]
