Respuesta :
Answer:
[tex] P(\bar X <78)[/tex]
And we can use the following z score:
[tex] z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
And replacing we got using the normal standard distribution table or excel:
[tex]P(\bar X <78)= P(Z<\frac{78-76}{\frac{12}{\sqrt{36}}}) = P(Z<1)=0.8413[/tex]
C.0.8413
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 records of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(76,12)[/tex]
Where [tex]\mu=76[/tex] and [tex]\sigma=12[/tex]
Since the distribution 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 select a sample of size n=36, and we want to calculate this probability:
[tex] P(\bar X <78)[/tex]
And we can use the following z score:
[tex] z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
And replacing we got using the normal standard distribution table or excel:
[tex]P(\bar X <78)= P(Z<\frac{78-76}{\frac{12}{\sqrt{36}}}) = P(Z<1)=0.8413[/tex]
C.0.8413
Answer:
(C) 0.8413
Step-by-step explanation:
Test statistic (z) = (sample mean - population mean) ÷ (sd/√n)
sample mean = 78
population mean = 76
sd = 12
z = (78 - 76) ÷ (12/√36) = 2 ÷ 2 = 1
The cumulative area of the test statistic is the probability that Professor Elderman's class of 36 has a class average below 78. The cumulative area is 0.8413
Therefore, the probability is 0.8413
