jane needs $20 to buy a radio. she already saved $15. what percent of the cost of the radio has she already saved

A normally distributed random variable with mean 4.5 and standard deviation 7.6 is sampled to get two independent values, X1 and

mr Goodwill [35]

Answer:

Bias for the estimator = -0.56

Mean Square Error for the estimator = 6.6311

Step-by-step explanation:

Given - A normally distributed random variable with mean 4.5 and standard deviation 7.6 is sampled to get two independent values, X1 and X2. The mean is estimated using the formula (3X1 + 4X2)/8.

To find - Determine the bias and the mean squared error for this estimator of the mean.

Proof -

Let us denote

X be a random variable such that X ~ N(mean = 4.5, SD = 7.6)

Now,

An estimate of mean, μ is suggested as

$\mu = \frac{3X_{1} + 4X_{2} }{8}$

Now

Bias for the estimator = E(μ bar) - μ

= $E( \frac{3X_{1} + 4X_{2} }{8}) - 4.5$

= $\frac{3E(X_{1}) + 4E(X_{2})}{8} - 4.5$

= $\frac{3(4.5) + 4(4.5)}{8} - 4.5$

= $\frac{13.5 + 18}{8} - 4.5$

= $\frac{31.5}{8} - 4.5$

= 3.9375 - 4.5

= - 0.5625 ≈ -0.56

∴ we get

Bias for the estimator = -0.56

Now,

Mean Square Error for the estimator = E[(μ bar - μ)²]

= Var(μ bar) + [Bias(μ bar, μ)]²

= $Var( \frac{3X_{1} + 4X_{2} }{8}) + 0.3136$

= $\frac{1}{64} Var( {3X_{1} + 4X_{2} }) + 0.3136$

= $\frac{1}{64} ( [{3Var(X_{1}) + 4Var(X_{2})] }) + 0.3136$

= $\frac{1}{64} [{3(57.76) + 4(57.76)}] } + 0.3136$

= $\frac{1}{64} [7(57.76)}] } + 0.3136$

= $\frac{1}{64} [404.32] } + 0.3136$

= $6.3175 + 0.3136$

= 6.6311

∴ we get

Mean Square Error for the estimator = 6.6311

6 0

3 years ago

9. The 10,000 seat stadium is 89% full. How many people are watching the game inside the stadium? with solution pls.

Ksivusya [100]

Answer:1100

Explanation:89% of 10,000 is 8900 subtract 8900 from 10,000 bam you got 1100

your welcome

7 0

2 years ago

Read 2 more answers

Test scores of the student in a school are normally distributed mean 85 standard deviation 3 points. What's the probability that

Mrrafil [7]

Answer:

The probability that a random selected student score is greater than 76 is $\\ P(x>76) = 0.99865$ .

Step-by-step explanation:

The Normally distributed data are described by the normal distribution. This distribution is determined by two parameters, the population mean $\\ \mu$ and the population standard deviation $\\ \sigma$ .

To determine probabilities for the normal distribution, we can use the standard normal distribution, whose parameters' values are $\\ \mu = 0$ and $\\ \sigma = 1$ . However, we need to "transform" the raw score, in this case x = 76, to a z-score. To achieve this we use the next formula:

$\\ z = \frac{x - \mu}{\sigma}$ [1]

And for the latter, we have all the required information to obtain z. With this, we obtain a value that represent the distance from the population mean in standard deviations units.

<h3>The probability that a randomly selected student score is greater than 76</h3>

To obtain this probability, we can proceed as follows:

First: obtain the z-score for the raw score x = 76.

We know that:

$\\ \mu = 85$

$\\ \sigma = 3$

$\\ x = 76$

From equation [1], we have:

$\\ z = \frac{76 - 85}{3}$

Then

$\\ z = \frac{-9}{3}$

$\\ z = -3$

Second: Interpretation of the previous result.

In this case, the value is three (3) standard deviations below the population mean. In other words, the standard value for x = 76 is z = -3. So, we need to find P(x>76) or P(x>-3).

With this value of $\\ z = -3$ , we can obtain this probability consulting the cumulative standard normal distribution, available in any Statistics book or on the internet.

Third: Determination of the probability P(x>76) or P(x>-3).

Most of the time, the values for the cumulative standard normal distribution are for positive values of z. Fortunately, since the normal distributions are symmetrical, we can find the probability of a negative z having into account that (for this case):

$\\ P(z>-3) = 1 - P(z>3) = P(z$

Then

Consulting a cumulative standard normal table, we have that the cumulative probability for a value below than three (3) standard deviations is:

$\\ P(z$

Thus, "the probability that a random selected student score is greater than 76" for this case (that is, $\\ \mu = 85$ and $\\ \sigma = 3$ ) is $\\ P(x>76) = P(z>-3) = P(z$ .