Question

A normally distributed population has mean 57,800 and standard deviation 750. Find the probability that a single randomly selected element X of the population is between 57,000 and 58,000. Find the mean and standard deviation of X - for samples of size 100. Find the probability that the mean of a sample of size 100 drawn from this population is between 57,000 and 58,000.

Answer 1

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.; $\mu$ = 57,800  and  $\sigma$ = 750.

Let X = randomly selected element of the population

The z probability is given by;

           Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)  

(a) So, P(57,000 <= X <= 58,000) = P(X <= 58,000) - P(X < 57,000)

P(X <= 58,000) = P( $\frac{X-\mu}{\sigma}$ <= $\frac{58000-57800}{750}$ ) = P(Z <= 0.27) = 0.60642

P(X < 57000) = P( $\frac{X-\mu}{\sigma}$ < $\frac{57000-57800}{750}$ ) = 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 = $\frac{\sigma}{\sqrt{n} }$ = $\frac{750}{\sqrt{100} }$ = 75

The z probability is given by;

           Z = $\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ 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( $\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }$ <= $\frac{58000-57800}{\frac{750}{\sqrt{100} } }$ ) = P(Z <= 2.67) = 0.99621

P(X < 57000) = P( $\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }$ < $\frac{57000-57800}{\frac{750}{\sqrt{100} } }$ ) = 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 .

Answer 2

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 $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

• 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 $\mu$ and the standard deviation is $s = \frac{\sigma}{\sqrt{n}}$.

In this problem:

• Mean of 57800, thus $\mu = 57800$.
• Standard deviation of 750, thus $\sigma = 750$.

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:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{58000 - 57800}{750}$

$Z = 0.27$

$Z = 0.27$ has a p-value of 0.6064.

X = 57000:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{57000 - 57800}{750}$

$Z = -1.07$

$Z = -1.07$ 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, $n = 100$, and then:

$s = \frac{750}{\sqrt{100}} = 75$

For samples of size 100, the mean is of 57800 and the standard deviation is 75.

Then, the probability is:

X = 58000:

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{58000 - 57800}{75}$

$Z = 2.7$

$Z = 2.7$ has a p-value of 0.9965.

X = 57000:

$Z = \frac{X - \mu}{s}$

$Z = \frac{57000 - 57800}{75}$

$Z = -10.7$

$Z = -10.7$ 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