Ask question

Ask question

All categories

3 years ago

12

If you have 8 marbles in the bag (5 blue and 3 green) and you draw 2 with replacement, what is the theoretical probability that

both marbles will be green?
a.
14%
c.
23%
b.
20%
d.
39%

1 answer:

Readme [11.4K]3 years ago

3 0

Answer:

d.39%

Step-by-step explanation:

Loki told me

But really, 39% should be correct

You might be interested in

Complete the following two-way frequency table.

Lera25 [3.4K]

Answer:

Step-by-step explanation:

Number of candies with Forest = 12

Candies containing coconut and chocolate both = Number common in coconut and the chocolate = 3

Candies which do not contain coconut but contain the chocolate = 6

Candies which contain the coconut but do not contain the chocolate = 1

Candies which neither contain the chocolate nor coconut = 2

From the given Venn diagram,

Contain coconut Do not contain coconut

Contain chocolate 3 6

Do not contain chocolate 1 2

6 0

3 years ago

Mr falcons pizza shop offers three sizes of pizza the small medium and large they have diameters of 8 inches 10 inches and 12 in

yuradex [85]

The circumference = 2×pi×radius
the small = 2×3.14×4 = 25.12 inches
the medium = 2×3.14×5 = 31.4 inches
the large = 2×3.14×6 = 37.68 inches

6 0

3 years ago

the following two triangles are similar. Find the missing side length represented by x. Please help i am almost done.

lbvjy [14]

Answer:

7 inches

Step-by-step explanation:

First, you make an imbalanced equation: 22.75/x=13/4. Then you cross multiply: 22.75 * 4= 91. Then divide 91/13 = 7.

5 0

3 years ago

The quality control manager of a chemical company randomly sampled twenty 100-pound bags of fertilizer to estimate the variance

lisabon 2012 [21]

Answer:

The 95% confidence interval for the variance in the pounds of impurities would be $3.829 \leq \sigma^2 \leq 14.121$ .

Step-by-step explanation:

1) Data given and notation

$s^2 =6.62$ represent the sample variance

s=2.573 represent the sample standard deviation

$\bar x$ represent the sample mean

n=20 the sample size

Confidence=95% or 0.95

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population mean or variance lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

The Chi square distribution is the distribution of the sum of squared standard normal deviates .

2) Calculating the confidence interval

The confidence interval for the population variance is given by the following formula:

$\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}$

On this case the sample variance is given and for the sample deviation is just the square root of the sample variance.

The next step would be calculate the critical values. First we need to calculate the degrees of freedom given by:

$df=n-1=20-1=19$

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical values.

The excel commands would be: "=CHISQ.INV(0.025,19)" "=CHISQ.INV(0.975,19)". so for this case the critical values are:

$\chi^2_{\alpha/2}=32.852$

$\chi^2_{1- \alpha/2}=8.907$

And replacing into the formula for the interval we got:

$\frac{(19)(6.62)}{32.852} \leq \sigma^2 \frac{(19)(6.62)}{8.907}$

$3.829 \leq \sigma^2 \leq 14.121$

So the 95% confidence interval for the variance in the pounds of impurities would be $3.829 \leq \sigma^2 \leq 14.121$ .

4 0

3 years ago

Let X1 and X2 be independent random variables with mean μand variance σ².

My name is Ann [436]

Answer:

a) $E(\hat \theta_1) =\frac{1}{2} [E(X_1) +E(X_2)]= \frac{1}{2} [\mu + \mu] = \mu$

So then we conclude that $\hat \theta_1$ is an unbiased estimator of $\mu$

$E(\hat \theta_2) =\frac{1}{4} [E(X_1) +3E(X_2)]= \frac{1}{4} [\mu + 3\mu] = \mu$

So then we conclude that $\hat \theta_2$ is an unbiased estimator of $\mu$

b) $Var(\hat \theta_1) =\frac{1}{4} [\sigma^2 + \sigma^2 ] =\frac{\sigma^2}{2}$

$Var(\hat \theta_2) =\frac{1}{16} [\sigma^2 + 9\sigma^2 ] =\frac{5\sigma^2}{8}$

Step-by-step explanation:

For this case we know that we have two random variables:

$X_1 , X_2$ both with mean $\mu = \mu$ and variance $\sigma^2$

And we define the following estimators:

$\hat \theta_1 = \frac{X_1 + X_2}{2}$

$\hat \theta_2 = \frac{X_1 + 3X_2}{4}$

Part a

In order to see if both estimators are unbiased we need to proof if the expected value of the estimators are equal to the real value of the parameter:

$E(\hat \theta_i) = \mu , i = 1,2$

So let's find the expected values for each estimator:

$E(\hat \theta_1) = E(\frac{X_1 +X_2}{2})$

Using properties of expected value we have this:

$E(\hat \theta_1) =\frac{1}{2} [E(X_1) +E(X_2)]= \frac{1}{2} [\mu + \mu] = \mu$

So then we conclude that $\hat \theta_1$ is an unbiased estimator of $\mu$

For the second estimator we have:

$E(\hat \theta_2) = E(\frac{X_1 + 3X_2}{4})$

Using properties of expected value we have this:

$E(\hat \theta_2) =\frac{1}{4} [E(X_1) +3E(X_2)]= \frac{1}{4} [\mu + 3\mu] = \mu$

So then we conclude that $\hat \theta_2$ is an unbiased estimator of $\mu$

Part b

For the variance we need to remember this property: If a is a constant and X a random variable then:

$Var(aX) = a^2 Var(X)$

For the first estimator we have:

$Var(\hat \theta_1) = Var(\frac{X_1 +X_2}{2})$

$Var(\hat \theta_1) =\frac{1}{4} Var(X_1 +X_2)=\frac{1}{4} [Var(X_1) + Var(X_2) + 2 Cov (X_1 , X_2)]$

Since both random variables are independent we know that $Cov(X_1, X_2 ) = 0$ so then we have:

$Var(\hat \theta_1) =\frac{1}{4} [\sigma^2 + \sigma^2 ] =\frac{\sigma^2}{2}$

For the second estimator we have:

$Var(\hat \theta_2) = Var(\frac{X_1 +3X_2}{4})$

$Var(\hat \theta_2) =\frac{1}{16} Var(X_1 +3X_2)=\frac{1}{4} [Var(X_1) + Var(3X_2) + 2 Cov (X_1 , 3X_2)]$

Since both random variables are independent we know that $Cov(X_1, X_2 ) = 0$ so then we have:

$Var(\hat \theta_2) =\frac{1}{16} [\sigma^2 + 9\sigma^2 ] =\frac{5\sigma^2}{8}$

7 0

3 years ago