Respuesta :
Answer:
(a) Probability that a single randomly selected element X of the population is between 57,000 and 58,000 = 0.46411
(b) Probability that the mean of a sample of size 100 drawn from this population is between 57,000 and 58,000 = 0.99621
Step-by-step explanation:
We are given that a normally distributed population has mean 57,800 and standard deviation 75, i.e.; [tex]\mu[/tex] = 57,800 and [tex]\sigma[/tex] = 750.
Let X = randomly selected element of the population
The z probability is given by;
Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ N(0,1)
(a) So, P(57,000 <= X <= 58,000) = P(X <= 58,000) - P(X < 57,000)
P(X <= 58,000) = P( [tex]\frac{X-\mu}{\sigma}[/tex] <= [tex]\frac{58000-57800}{750}[/tex] ) = P(Z <= 0.27) = 0.60642
P(X < 57000) = P( [tex]\frac{X-\mu}{\sigma}[/tex] < [tex]\frac{57000-57800}{750}[/tex] ) = P(Z < -1.07) = 1 - P(Z <= 1.07)
= 1 - 0.85769 = 0.14231
Therefore, P(31 < X < 40) = 0.60642 - 0.14231 = 0.46411 .
(b) Now, we are given sample of size, n = 100
So, Mean of X, X bar = 57,800 same as before
But standard deviation of X, s = [tex]\frac{\sigma}{\sqrt{n} }[/tex] = [tex]\frac{750}{\sqrt{100} }[/tex] = 75
The z probability is given by;
Z = [tex]\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] ~ N(0,1)
Now, probability that the mean of a sample of size 100 drawn from this population is between 57,000 and 58,000 = P(57,000 < X bar < 58,000)
P(57,000 <= X bar <= 58,000) = P(X bar <= 58,000) - P(X bar < 57,000)
P(X bar <= 58,000) = P( [tex]\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] <= [tex]\frac{58000-57800}{\frac{750}{\sqrt{100} } }[/tex] ) = P(Z <= 2.67) = 0.99621
P(X < 57000) = P( [tex]\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] < [tex]\frac{57000-57800}{\frac{750}{\sqrt{100} } }[/tex] ) = P(Z < -10.67) = P(Z > 10.67)
This probability is that much small that it is very close to 0
Therefore, P(57,000 < X bar < 58,000) = 0.99621 - 0 = 0.99621 .
Using the normal distribution and the central limit theorem, it is found that:
- There is a 0.4641 = 46.41% probability that a single randomly selected element X of the population is between 57,000 and 58,000.
- For samples of size 100, the mean is of 57800 and the standard deviation is 75.
- There is a 0.9965 = 99.65% probability that the mean of a sample of size 100 drawn from this population is between 57,000 and 58,000.
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.
- By the Central Limit Theorem, for sampling distributions of samples of size n, the mean is [tex]\mu[/tex] and the standard deviation is [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
In this problem:
- Mean of 57800, thus [tex]\mu = 57800[/tex].
- Standard deviation of 750, thus [tex]\sigma = 750[/tex].
The probability that a single randomly selected element X of the population is between 57,000 and 58,000 is the p-value of Z when X = 58000 subtracted by the p-value of Z when X = 57000, thus:
X = 58000:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{58000 - 57800}{750}[/tex]
[tex]Z = 0.27[/tex]
[tex]Z = 0.27[/tex] has a p-value of 0.6064.
X = 57000:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{57000 - 57800}{750}[/tex]
[tex]Z = -1.07[/tex]
[tex]Z = -1.07[/tex] has a p-value of 0.1423.
0.6064 - 0.1423 = 0.4641.
0.4641 = 46.41% probability that a single randomly selected element X of the population is between 57,000 and 58,000.
For samples of size 100, [tex]n = 100[/tex], and then:
[tex]s = \frac{750}{\sqrt{100}} = 75[/tex]
For samples of size 100, the mean is of 57800 and the standard deviation is 75.
Then, the probability is:
X = 58000:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{58000 - 57800}{75}[/tex]
[tex]Z = 2.7[/tex]
[tex]Z = 2.7[/tex] has a p-value of 0.9965.
X = 57000:
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{57000 - 57800}{75}[/tex]
[tex]Z = -10.7[/tex]
[tex]Z = -10.7[/tex] has a p-value of 0.
0.9965 - 0 = 0.9965.
0.9965 = 99.65% probability that the mean of a sample of size 100 drawn from this population is between 57,000 and 58,000.
A similar problem is given at https://brainly.com/question/24663213
