12x+7<-11 and 5x-8>= 40

Answer:

A. The interval will be narrower if 15 men are used in the sample.

Step-by-step explanation:

Hello!

When all other things remain the same, which of the following statements about the width of the interval is correct?

A. The interval will be narrower if 15 men are used in the sample.

B. The interval will be wider if 15 men are used in the sample.

C. The interval will be narrower if 5 men are used in the sample.

D. The interval will be narrower if the level is increased to 99% confidence.

E. The interval will be wider if the level is decreased to 90% confidence.

Consider that the variable of interest "Xd: Difference between the peak power of a cyclist before training and after training" has a normal distribution. To construct the confidence interval for the population mean of the difference you have to use a pooled t-test.

The general structure for the CI is "point estimate"±" margin fo error"

Any modification to the sample size, sample variance and/or the confidence level affect the length of the interval (amplitude) and the margin of error (semiamplitude)

The margin of error of the interval is:

d= [tex]t_{n-1;1-\alpha /2}[/tex] * (Sd/n)

1) The sample size changes, all other terms of the interval stay the same.

As you can see the margin of error and the sample size (n) have an indirect relationship. This means, that when the sample size increases, the semiamplitude decreases, and when the sample size decreases, the semiamplitude increases.

↓d= [tex]t_{n-1;1-\alpha /2}[/tex] * (Sd/↑n)

↑d= [tex]t_{n-1;1-\alpha /2}[/tex] * (Sd/↓n)

Correct option: A. The interval will be narrower if 15 men are used in the sample.

2) The confidence level has a direct relationship with the semiamplitude of the interval, this means that when the confidence level increases, so do the semiamplitude, and if the level decreases, so do the semiamplitude:

↓d= ↓[tex]t_{n-1;1-\alpha /2}[/tex] * (Sd/n)

↑d= ↑[tex]t_{n-1;1-\alpha /2}[/tex] * (Sd/n)

I hope it helps!

Answer:

The interval will be narrower if 15 men are used in the sample.

Step-by-step explanation:

Answer:

25.5 cm²

Step-by-step explanation:

5.5 cm × 20 cm = 25.5 cm²

Answer:i think the answer is 110 but it would be more helpful if i knew what kind of shape it is

Step-by-step explanation:

to find the area of a square or rectangle(assuming this is a square or rectangle) you multiply the base by the height

Answer:

y = sin(t) is a solution to the differential equation

(dy/dt)² = 1 - y²

Step-by-step explanation:

Given (dy/dt)² = 1 - y²

Suppose y = sin(t) is a solution, then it satisfies the differential equation.

That is

[d(sin(t))]² = 1 - y²

Let y = sin(t)

dy/dt = d(sin(t)) = cos(t)

(dy/dt)² = cos²t

But cos²t + sin²t = 1

=> 1 - sin²t = cos²t

So

(dy/dt)² = 1 - sin²t

Since sin²t = (sint)² = y²,

we have

(dy/dt)² = 1 - y²

as required.

The differential equation becomes [tex](\frac{dy}{dx} )^2 = 1-y^2 (Proved)[/tex]

Given the function;

Take the differential of the function

Square both sides of the equation to have:

[tex](\frac{dy}{dx} )^2 = (cost)^2[/tex]

Recall from trigonometry identity that [tex]sin^2t + cos^2t = 1[/tex]

Hence, [tex]cos^2t = 1- sin^2t[/tex]

Replace into the differential expression to have:

[tex](\frac{dy}{dx} )^2 = 1-sin^2t[/tex]

Recall that y = sin(t). On replacing, the differential equation becomes:

[tex](\frac{dy}{dx} )^2 = 1-y^2 (Proved)[/tex]

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Answer:

50

Step-by-step explanation:

Answer:

the answer is 200

Step-by-step explanation:

Answer:

[tex] ME = 1.833 * \frac{4.367}{\sqrt{10}}= 2.531[/tex]

Step-by-step explanation:

Previous concepts

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 means 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.

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".

[tex]\bar X[/tex] represent the sample mean for the sample  

[tex]\mu[/tex] population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size  

Solution to the problem

The confidence interval for the mean is given by the following formula:

[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex]   (1)

In order to calculate the mean and the sample deviation we can use the following formulas:  

[tex]\bar X= \sum_{i=1}^n \frac{x_i}{n}[/tex] (2)  

[tex]s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}}[/tex] (3)  

The mean calculated for this case is [tex]\bar X=24.8[/tex]

The sample deviation calculated [tex]s=4.367[/tex]

In order to calculate the critical value [tex]t_{\alpha/2}[/tex] we need to find first the degrees of freedom, given by:

[tex]df=n-1=10-1=9[/tex]

Since the Confidence is 0.90 or 90%, the value of [tex]\alpha=0.1[/tex] and [tex]\alpha/2 =0.005[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.05,9)".And we see that [tex]t_{\alpha/2}=1.833[/tex]

And the margin of error would be given by:

[tex] ME = 1.833 * \frac{4.367}{\sqrt{10}}= 2.531[/tex]

The amount the florist earns for each delivery, with 'x' being the miles away from the flower shop, can be modeled with the equation: Earnings = (7.5 + 0.75x). This represents the fixed net income of $7.5 and $0.75 per mile after paying the driver.

The florist charges ​$12.00 for delivery and an additional ​$1.50 per mile from the flower shop. However, the florist also has costs to cover, namely $0.75 per mile to pay the driver, and ​$4.50 for gas per delivery. The net earning per delivery, with 'x' representing the number of miles a delivery location is from the flower shop, can be modeled by the following algebraic expression: Earnings = (12 + 1.5x) - (0.75x + 4.5).

This actually simplifies to: Earnings = (7.5 + 0.75x). The 7.5 is the fixed net income for each delivery (gross earnings minus the gasoline cost) and 0.75x is the per-mile net income after the driver is paid.

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Final answer:

To find out how many people will be in Houston, Texas in 20 years, we can use the doubling time of 35 years. Currently, the population is 2.4 million. Using the formula New Population = Current Population x 2^(Number of Doubling Cycles), the new population will be approximately 4.8 million. Option A .

Explanation:

To find out how many people will be in Houston, Texas in 20 years, we can use the doubling time of 35 years. Currently, the population is 2.4 million. Since the city doubles in size every 35 years, in 20 years it will go through 20/35 of a doubling cycle.

To calculate how many people there will be, we use the formula:

New Population = Current Population x 2^(Number of Doubling Cycles)

Plugging in the values, we have:

New Population = 2.4 million x 2^(20/35)

Using a calculator, we find that the new population will be approximately 4.8 million. Therefore, the answer is 4.8 million.

Answer:

22

Step-by-step explanation:

PEMDAS

4^2 = 16

16 - 5 = 11

11 x 2 = 22

Answer:

22

Step-by-step explanation:

2 (c^2 -5)

Let c=4

2 (4^2 -5)

PEMDAS

Parentheses first,

(4^2 -5)

Exponents

16-5 =11

Replace into expression

2(11)

22

Answer:

1.045

Step-by-step explanation:

Answer:

3/2

Step-by-step explanation:

Answer:

-10; no

Step-by-step explanation:

Final answer:

The inner product of the vectors (-4,9,8) and (3,2,-2) is -10. Since the inner product is not zero, the vectors are not perpendicular. Therefore, the correct answer is: -10; no.

Explanation:

The inner product (also known as the dot product) of two vectors (-4,9,8) and (3,2,-2) is calculated by multiplying the corresponding components of the two vectors and summing the result:

Inner product = (-4)×3 + 9×2 + 8×(-2)
= -12 + 18 - 16
= -10

To determine if the vectors are perpendicular, we check if their inner product is zero. Since the inner product in this case is -10, not zero, the vectors are not perpendicular.

Answer:

0.625

Step-by-step explanation:

Answer:

(A) Type I error is convicting an innocent person.

Step-by-step explanation:

When someone is on trial for suspicion of committing a crime, the hypotheses are:

[tex]H_0[/tex] : innocent

[tex]H_A[/tex] : guilty

From the given options, A Type I error is convicting an innocent person. If the null hypothesis holds (i.e. a person is innocent) nut we still go ahead to convict the person, we have rejected a true null hypothesis.

Answer:

2.28% probability that the mean amount of their monthly cell phone bills is more than $60

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score 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. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

[tex]\mu = 58.90, \sigma = 3.64, n = 44, s = \frac{3.64}{\sqrt{44}} = 0.54875[/tex]

What is the probability that the mean amount of their monthly cell phone bills is more than $60?

This is 1 subtracted by the pvalue of Z when X = 60. So

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

By the Central Limit Theorem

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{60 - 58.90}{0.54875}[/tex]

[tex]Z = 2[/tex]

[tex]Z = 2[/tex] has a pvalue of 0.9772

1 - 0.9772 = 0.0228

2.28% probability that the mean amount of their monthly cell phone bills is more than $60

Answer:

A) probability that an arriving customer will wait in line is 67%

B)the probability that both windows are idle is 0.33

C) the average length of the waiting line is 1.33

D)it would not be possible to offer a reasonable service with only one window

Step-by-step explanation:

arrival rate: δ = 20 x 0.80 = 16 customers per hour

service rate: μ = 2 × (60/5) = 24 customers/hour

Utilization factor is given as;

Φ = δ/μ

So, Φ = 16/24 ≈ 0.67

A) the probability that an arriving customer will wait in line is;

16/24 x 100% ≈ 67%

B) probability that both windows are idle is;

P(x=0) = 1 - 0.67 = 0.33

C) The average number of customers in the post office will be;

L_s = Φ/(1 - Φ)

L_s = 0.67/(1 - 0.67)

L_s = 0.67/0.33

L_s ≈ 2 customers

Thus, the average length of the waiting line is;

L_w = L_s - Φ

L_w = 2 - 0.67

L_w = 1.33

D) this part demands that we find the utilization factor with only one window.

Thus;

arrival rate: δ = 20 x 0.80 = 16 customers per hour

And

service rate: μ = 1 × (60/5) = 12 customers/hour

Thus, Utilization factor = 16/12 = 1.33

Thus, it would not be possible to offer a reasonable service with only one window

Answer:
The equivalent system of first-order differential equations is:

[tex]\[ \begin{cases}u' = v \\v' = 8\sin(3t) - 6.5v + 1.5u\end{cases} \][/tex]

with the initial conditions:

[tex]\[ \begin{cases}u(1) = -3 \\v(1) = -1.5\end{cases} \][/tex]


To convert the given second-order differential equation into an equivalent system of first-order equations, we follow these steps:

1. Identifying the original second-order differential equation:

  u'' + 6.5u' - 1.5u = 8sin(3t)

2. Then we introduce a new variable v to represent the first derivative of u:

  v = u'

3. Now ,we write the second-order equation as a system of first-order equations:

  The original equation is:

  u'' + 6.5u' - 1.5u = 8sin(3t)

  Using the new variable v = u' , the second derivative u'' can be written as v'. Therefore, we have:

  v' + 6.5v - 1.5u = 8sin(3t)

4. Formulating the system of first-order equations:

  The first equation comes directly from the definition of v:

  u' = v

  The second equation comes from the rewritten second-order equation:

  v' = 8sin(3t) - 6.5v + 1.5u

5. Let us include the initial conditions:

  The initial conditions provided are:

  u(1) = -3

  u'(1) = -1.5

  Since v = u', this translates to:

  v(1) = -1.5

6. Writing the system with initial conditions:

  The system of first-order equations is:

[tex]\[ \begin{cases} u' = v \\ v' = 8\sin(3t) - 6.5v + 1.5u \end{cases} \][/tex]

  With the initial conditions:

 [tex]\[ \begin{cases} u(1) = -3 \\ v(1) = -1.5 \end{cases} \][/tex]

Answer:

43/91 or 47%

Step-by-step explanation:

Answer:

The critical value of t for 99% confidence interval is 2.898.

Step-by-step explanation:

The complete question is:

A biologist is trying to determine the average age of a local forest. She cuts down 18 randomly  selected trees and counts the tree rings. They find the average number of tree rings to be 83 with a  variance of 320. What is the critical value for the 99% confidence interval?

The population variance is not known and the sample size is too small. So a t-confidence interval will be used to estimate the population mean age of a local forest.

The (1 - α)% confidence interval for population mean is:

[tex]CI=\bar x\pm t_{\alpha/2, (n-1)}\times \frac{s}{\sqrt{n}}[/tex]

The information provided is:

n = 18

(1 - α)% = 99%

The degrees of freedom of the critical value of t is:

n - 1 = 18 - 1 = 17

Compute the critical value of t as follows:

[tex]t_{\alpha/2, (n-1)}=t_{0.01/2, (18-1)}=t_{0.005, 17}=2.898[/tex]

*Use a t-table.

Thus, the critical value of t for 99% confidence interval is 2.898.

Answer:

D.

Step-by-step explanation:

Rearrange

[tex]0.95x - 5.00 < 33.95[/tex]

[tex]0.95x<38.95[/tex]

[tex]x<\frac{38.95}{0.95}[/tex]

[tex]x<41[/tex]

Step-by-step explanation:

x²=a²+b²

x=√6²+12²

x=√180

x=3√2v

y²=16²+12²

y=√400

y=20

Answer: the answer for rafter 1 is 13.4 and the answer for rafter 2 is 20

Step-by-step explanation: I just know

Answer:

The slope is -1

Step-by-step explanation:

Let's find the slope between your two points.

(1,4);(6,−1)

(x1,y1)=(1,4)

(x2,y2)=(6,−1)

Use the slope formula:

m= y2−y1/x2−x1  = −1−4/6−1

= −5/5

= −1

Hope this is a good explanation :)

Answer:

Do the data from this shipment indicate statistical control: No

Step-by-step explanation:

Calculating the mean of the sample, we have;

Mean (x-bar) = sum of individual sample/number of sample

                     = (0+0+0+0.004+0.008+0.020+0.004+0+0+0.008)/10

                     = 0.044/10

                    = 0.0044

Calculating the lower control limit (LCL) using the formula;

LCL= (x-bar) - 3*√(x-bar(1-x-bar))/n

      = 0.0044 - 3*√(0.0044(1-0.0044))

       = 0.0044- (3*0.0042)

        = 0.0044 - 0.01256

        = -0.00816 ∠ 0

Calculating the upper control limit (UCL) using the formula;

UCL = (x-bar) + 3*√(x-bar(1-x-bar))/n

      = 0.0044 + 3*√(0.0044(1-0.0044))

       = 0.0044+ (3*0.0042)

        = 0.0044 + 0.01256

       =0.01696∠ 0

Do the data from this shipment indicate statistical control: No

Since the value 0.02 from the 6th shipment is greater than the upper control limit (0.01696), we can conclude that  the data from this shipment do not indicate statistical control.

The data from this shipment does not indicate statistical control.

Calculating the mean of the sample, we have;

Mean (x-bar) = sum of individual sample/number of sample

[tex]\frac{(0+0+0+0.004+0.008+0.020+0.004+0+0+0.008)}{10}\\=\frac{0.044}{10}\\=0.0044[/tex]

Calculating the lower control limit (LCL) using the formula;

LCL

= (x-bar) - 3*√(x-bar(1-x-bar))/n

[tex]= 0.0044 - 3*\sqrt{(0.0044(1-0.0044))}\\= 0.0044- (3*0.0042)\\= 0.0044 - 0.01256\\= -0.00816[/tex]

Calculating the upper control limit (UCL) using the formula;

UCL = (x-bar) + 3*√(x-bar(1-x-bar))/n

[tex]= 0.0044 + 3*\sqrt{(0.0044(1-0.0044))}\\= 0.0044+ (3*0.0042)\\= 0.0044 + 0.01256\\=0.01696[/tex]

Do the data from this shipment indicate statistical control:

Since the value 0.02 from the 6th shipment is greater than the upper control limit (0.01696), we can conclude that  the data from this shipment does not indicate statistical control.

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Answer: No, these results are not statistically significant because

p > 0.05

Step-by-step explanation:

The null hypothesis is

H0 : μ = 17

The alternative hypothesis is

H 1 : μ ≠ 17

where μ is the mean amount of cereal in each box.

The p value that he got is 0.1499. This is greater than alpha = 0.05 which is the given level of significance.

If the level of significance is lesser than the p value, we would accept the null hypothesis.

Therefore, the correct option is

No, these results are not statistically significant because p>0.05

Answer:

Step-by-step explanation:

The volume of a cuboid is the product of its dimensions:

  V = LWH = (3 ft)(3 ft)(2 ft)

  V = 18 ft³

The area is the sum of the areas of the faces. Since opposite faces have the same area, we can figure the area from ...

  A = 2(LW +H(L+W)) = 2((3 ft)(3 ft) +(2 ft)(3 ft +3 ft)) = 2(9 ft² +12 ft²)

  A = 42 ft²

Answer: The 95% confidence interval will be wider.

Step-by-step explanation:

Confidence interval for population proportion is written as

Sample proportion ± margin of error

margin of error = z score × √pq/n

The z score is determined by the confidence level. The z score for a confidence level of 95% is higher than the z score for a confidence level of 90%

This means that with all other things being equal, a 95% confidence level will give a higher margin of error compared to a 90% confidence level.

The higher the margin of error, the wider the confidence interval. Therefore,

The 95% confidence interval will be wider.

The correct statement is provided by Trevor: 'The 95% confidence interval will be wider.'

Given that,

If construct a 90% confidence interval for the population proportion,

And a 95% confidence interval for the population proportion,

Tim: 'The 90% confidence interval will be wider.'

Trevor: 'The 95% confidence interval will be wider.

Tracy: 'There is not enough information to tell.

Either interval could be wider.'

We have to determine,

Which of these intervals will be wider.

According to the question,

Three students provide their answers as follows:

Tim: 'The 90% confidence interval will be wider.

'Trevor: 'The 95% confidence interval will be wider.

'Tracy: 'There is not enough information to tell. Either interval could be wider.'

Therefore, Confidence interval for population proportion is written as

Sample proportion ± margin of error

Margin of error  [tex]= z score \times \frac{ \sqrt{pq}}{n}[/tex]

The z score for a confidence level of 95% is higher than the z score for a confidence level of 90%.

Other things being equal, a 95% confidence level will give a higher margin of error compared to a 90% confidence level.

The higher the margin of error, the wider the confidence interval.

Therefore, The 95% confidence interval will be wider.

Hence, The correct statement is provided by Trevor: 'The 95% confidence interval will be wider.'

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Answer:

86.65% probability that more than 70 workers will meet with an accident during the 1-yr period

Step-by-step explanation:

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

[tex]E(X) = np[/tex]

The standard deviation of the binomial distribution is:

[tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/tex]

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score 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. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that [tex]\mu = E(X)[/tex], [tex]\sigma = \sqrt{V(X)}[/tex].

In this problem, we have that:

[tex]p = 0.08, n = 1000[/tex]

So

[tex]\mu = E(X) = np = 1000*0.08 = 80[/tex]

[tex]\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{1000*0.08*0.92} = 8.58[/tex]

What is the probability that more than 70 workers will meet with an accident during the 1-yr period

Using continuity correction, this is [tex]P(X \geq 70 + 0.5) = P(X \geq 70.5)[/tex], which is 1 subtracted by the pvalue of Z when X = 70.5. So

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{70.5 - 80}{8.58}[/tex]

[tex]Z = -1.11[/tex]

[tex]Z = -1.11[/tex] has a pvalue of 0.1335

1 - 0.1335 = 0.8665

86.65% probability that more than 70 workers will meet with an accident during the 1-yr period

The probability that more than 70 workers will be involved in an accident is 0.8665

The given parameters are:

[tex]\mathbf{n = 1000}[/tex] --- population

[tex]\mathbf{p = 0.08}[/tex] --- the probability that a worker meets an accident

[tex]\mathbf{x = 70}[/tex] -- the number of workers

Start by calculating the mean and the standard deviation

[tex]\mathbf{\mu = np}[/tex] --- mean

So, we have:

[tex]\mathbf{\mu = 1000 \times 0.08}[/tex]

[tex]\mathbf{\mu = 80}[/tex]

[tex]\mathbf{\sigma = \sqrt{\mu(1 - p)}}[/tex]

So, we have:

[tex]\mathbf{\sigma = \sqrt{80 \times (1 - 0.08)}}[/tex]

[tex]\mathbf{\sigma = \sqrt{73.6}}[/tex]

[tex]\mathbf{\sigma = 8.58}[/tex]

The probability is then represented as

[tex]\mathbf{P(x > 70) = P(x > 70.5)}[/tex] ---- By continuity correction

Calculate the z-score for x = 70.5

[tex]\mathbf{z = \frac{x - \mu}{\sigma}}[/tex]

So, we have:

[tex]\mathbf{z = \frac{70.5 - 80}{8.58}}[/tex]

[tex]\mathbf{z = -1.11}[/tex]

So, we have:

[tex]\mathbf{P(x > 70) = P(z > -1.11)}[/tex]

Using z-scores of probabilities, we have:  

[tex]\mathbf{P(x > 70) = 0.8665}[/tex]

Hence, the probability that more than 70 workers will be involved in an accident is 0.8665

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Answer:14

Step-by-step explanation:

Answer:

5 faces

Step-by-step explanation:

4 triangular, 1 square

Answer:

12.75

Step-by-step explanation:

Answer:

1. Mean is 1.75

2. The variance is 1.6042

3.

The distribution function is:

Z           Z/K

0           21/128

1             5/16

2         33/128

3          1/6

4           29/384

5             1/48

6          1/384

Step-by-step explanation:

The mean of Z is given as:

E(Z) =Σ6, k=0 Kp (Z = k)

Σ6,k=0 K 1/6 Σ6, n=k (n k) (1/2)^n

=( 0(21/128) + 1(5/16) + 2( 33/128) + 3 (1/6) + 4 (29/384) + 5 (1/48) + 6 (1/384))

=7/4

=1.75

Thus, the mean Z is 1.75

The variance of Z is given as:

Var (Z) = E (Z^2) - (E (Z)) ^2

Therefore,

E(Z^2) = Σ 6, k=0 K^2P ( Z=K)

= ( 0(21/128 + 1(5/16) + 4(33/128) + 9(1/6) + 16(29/384) + 25(1/48) + 36(1/384))

=14/3

Var (Z) = 14/7 - (7/4)^2

= 14/7 - 49/16

=77/48

=1.6042

Thus, the variance is 1.6042

The probability of mass function is given as:

P(Z=k) = 1/6 Σ 6, n=k (n  k) (1/2)^n

The distribution function is

Z           Z/K

0           21/128

1             5/16

2         33/128

3          1/6

4           29/384

5             1/48

6          1/384