According to an exit poll for an​ election, 55.6​% of the sample size of 836 reported voting for a specific candidate. Is this enough evidence to predict who​ won? Test that the population proportion who voted for this candidate was 0.50 against the alternative that it differed from 0.50.

Report the test statistic and P-value and interpret the latter.

Final answer:

To test whether the percentage of correctly guessed penalty shots is less than 50% in World Cup Soccer, we can use a one-sample proportion test. Using a significance level of 0.05, we find that the test statistic is -2.09. Comparing this to the critical value from the standard normal distribution (-1.645), we reject the null hypothesis and conclude that there is evidence to suggest that the percentage of correctly guessed penalty shots is less than 50%.

Explanation:

To test whether there is evidence that the percentage of correctly guessed penalty shots is less than 50% in World Cup Soccer, we can use a one-sample proportion test. We will assume that the null hypothesis is true and that the goalkeeper's correct guesses are no better than random chance. The alternative hypothesis would be that the goalkeeper's correct guesses are significantly less than 50%. Using a significance level of 0.05, we can calculate the test statistic and compare it to the critical value from the standard normal distribution.

Null hypothesis (H0): The percentage of correctly guessed penalty shots is 50%.

Alternative hypothesis (Ha): The percentage of correctly guessed penalty shots is less than 50%.

Test statistic: We can use the z-test statistic since the sample size is large enough. The formula for the z-test statistic is z = (p - P0) / SE, where p is the sample proportion, P0 is the hypothesized proportion, and SE is the standard error. In this case, since the standard error is given as 0.043, we can plug in the values to calculate the test statistic.

Calculate the z-test statistic: z = (0.41 - 0.5) / 0.043 = -2.09

Find the critical value: Since our alternative hypothesis is that the percentage is less than 50%, we will use a one-tailed test. With a significance level of 0.05, the critical value from the standard normal distribution is -1.645.

Compare the test statistic to the critical value: Since the test statistic (-2.09) is less than the critical value (-1.645), we can reject the null hypothesis. There is evidence to suggest that the percentage of correctly guessed penalty shots is less than 50% in World Cup Soccer.

Answer:

(a) H1: μ < 530. Reject H1 if tcalc > –1.753

(b) t calc = –1.6. There is not enough evidence to reject the manufacturer’s claim.

Step-by-step explanation:

We are given that the manufacturer of an airport baggage scanning machine claims it can handle an average of 530 bags per hour.

A sample of 16 randomly chosen hours with a mean of 510 and a standard deviation of 50 is given.

Let [tex]\mu[/tex] = average bags an airport baggage scanning machine can handle

So, Null Hypothesis, [tex]H_0[/tex] : [tex]\mu \geq[/tex] 530 bags     {means that an airport baggage scanning machine can handle an average of more than or equal to 530 bags per hour}

Alternate Hypothesis, [tex]H_A[/tex] : [tex]\mu[/tex] < 530 bags     {means that an airport baggage scanning machine can handle an average of less than 530 bags per hour}

The test statistics that would be used here One-sample t test statistics as we don't know about the population standard deviation;

                         T.S. =  [tex]\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }[/tex]  ~ [tex]t_n_-_1[/tex]

where, [tex]\bar X[/tex] = sample mean = 510

             s = sample standard deviation = 50

            n = sample of hours = 16

So, test statistics  =  [tex]\frac{510-530}{\frac{50}{\sqrt{16} } }[/tex]  ~ [tex]t_1_5[/tex]

                              =  -1.60

The value of t test statistics is -1.60.

Now, at 0.05 significance level the t table gives critical value of -1.753 at 15 degree of freedom for left-tailed test. Since our test statistics is more than the critical values of t as -1.60 > -1.753, so we have insufficient evidence to reject our null hypothesis as it will not in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that an airport baggage scanning machine can handle an average of more than or equal to 530 bags per hour.

Answer:

The theoretical probability of landing on 2 heads, when 10 coins are tossed is 0.0439 or 4.39%.

Step-by-step explanation:

Number of coins that fell on the floor = 10

Number of coins that landed on heads = 2

We have to find the theoretical probability of getting 2 coins landing of heads when 10 coins are tossed.

Notice that there are only 2 possible outcomes: Either that coin will land on head or it won't. Landing of each coin is independent of the others coins. Probability of each coin landing on head is constant i.e. 0.5 or 1/2. Number of trials, i.e. the number of times the experiment will be done is fixed, which is 10.

All the 4 conditions for an experiment to be considered a Binomial Experiment are satisfied. So we will use Binomial Probability to solve this problem.

Probability of success = Probability of coin landing on head = 0.5

Number of trials = n = 10

Number of success = r = 2

The formula for Binomial Probability is:

[tex]P(X = x) =^{n}C_{r}(p)^{r}(1-p)^{n-r}[/tex]

Using the values, we get:

[tex]P(X=2)=^{10}C_{2}(0.5)^2(0.5)^8=0.0439[/tex]

Thus, the theoretical probability of landing on 2 heads, when 10 coins are tossed is 0.0439 or 4.39%.

Answer:

A

Step-by-step explanation:

Cause I know

Answer: C!!

Step-by-step explanation:

USA Test Prep told me! :)

Answer:

[tex]\frac{33}{36}[/tex]

Step-by-step explanation:

Combinations greater than a 10

5 - 5

5 - 6

6 - 5

There are total 36 combinations (6 * 6).

3 of these combinations are higher than 10.

So 36 - 3 combinations are less than 10.

Answer:

im lost good luck

Step-by-step explanation:

Answer:

10

3.4641

2.8965  

12.8965  

Step-by-step explanation:

Given: 10, 8, 12, 15, 13, 11, 6, 5  

c = 95%

a. The point estimate of the population mean is the sample mean. The mean is the sum of all values divided by the number of values:

x = 10 + 8 + 12 + 15 + 13 + 11 + 6 + 5 /8

  = 80/8

  = 10

b. The point estimate of the population standard deviation is the sample standard deviation. The variance is the sum of squared deviations from the mean divided by n - 1. The standard deviation is the square root of the variance:  

s = /(10 – 10)^2 +.... + (5– 10)^2/8 – 1  

s = 3.4641

c. Determine the t-value by looking in the row starting with degrees of freedom df = n-1 = 8 –1 = 7 and in the column with [tex]\alpha[/tex] = (1 – c)/2 = 0.025 in table :  

t_[tex]\alpha[/tex]/2 = 2.365  

The margin of error is then:  

E = t_[tex]\alpha[/tex]/2 * s/√n

  = 2.365 x s 3.4641/ √8

  = 2.8965  

d. The confidence intent)] then becomes:  

7.1035 = 10 – 2.8965 = x – E <u<x +E= 10 + 2.8965 = 12.8965  

The point estimate of the population mean will be 10.

The point estimate of the population mean will be calculated thus:

= (10 + 8 + 12 + 15 + 13 + 11 + 11 + 6 + 5) / 8

= 80/8

= 10

Also, the margin of error will be:

= 2.365 × 3.4641/✓8

= 2.8965

In conclusion, the margin of error is 2.8965.

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Answer:

y=6x+7

Step-by-step explanation:

−6x+y=7

Step 1: Add 6x to both sides

y=6x+7

Answer:

y=6x+7

Step-by-step explanation:

whenever you solve for a letter you have to get it by itself. That means it has to be alone on the opposite side of the equation

move the -6x to the other side of the equation

that -6 changes to a  positive 6 because, your moving it to the other side of the equation

your left with y=6x+7

pls mark me brainliest

Answer:

[tex]z=1.28<\frac{a-490}{10}[/tex]

And if we solve for a we got

[tex]a=490 +1.28*10=502.8[/tex]

So the value of height that separates the bottom 90% of data from the top 10% is 502.8.  

Step-by-step explanation:

Previous concepts

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".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the weeknly amount of a population, and for this case we know the distribution for X is given by:

[tex]X \sim N(490,10)[/tex]  

Where [tex]\mu=490[/tex] and [tex]\sigma=10[/tex]

And the best way to solve this problem is using the normal standard distribution and the z score given by:

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

For this part we want to find a value a, such that we satisfy this condition:

[tex]P(X>a)=0.1[/tex]   (a)

[tex]P(X<a)=0.9[/tex]   (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.  

As we can see on the figure attached the z value that satisfy the condition with 0.9 of the area on the left and 0.1 of the area on the right it's z=1.28. On this case P(Z<1.28)=0.9 and P(z>1.28)=0.1

If we use condition (b) from previous we have this:

[tex]P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.9[/tex]  

[tex]P(z<\frac{a-\mu}{\sigma})=0.9[/tex]

But we know which value of z satisfy the previous equation so then we can do this:

[tex]z=1.28<\frac{a-490}{10}[/tex]

And if we solve for a we got

[tex]a=490 +1.28*10=502.8[/tex]

So the value of height that separates the bottom 90% of data from the top 10% is 502.8.  

Final answer:

The company should budget approximately $502.80 for weekly repairs and maintenance to ensure that the probability of exceeding this amount is only 0.1.

Explanation:

We want to find how much should be budgeted for weekly repairs and maintenance so that the probability the budgeted amount will be exceeded in a given week is only 0.1.

Since the weekly amount of money spent is normally distributed with a mean of $490 and a standard deviation of $10, we can find the amount by looking up the z-score that corresponds to the 90th percentile

(since 100% - 10% = 90%) in a standard normal distribution table or using a calculator.

Let the z-score for the 90th percentile be denoted as z.

Looking up the standard normal distribution table or using a calculator, we find that z ≈ 1.28 for 0.9 cumulative probability.

We then use the z-score formula:

z = (X - mean) / standard deviation

Plugging in our z-score and the parameters, we can solve for X:

1.28 = (X - 490) / 10

X - 490 = 12.8

X = $502.80

Therefore, the company should budget approximately $502.80 for weekly repairs and maintenance to ensure that the probability of exceeding this amount is only 0.1.

Answer:

a)

USL = 6.2 inches

LSL = 5.8 inches

b) Cp = 1.33

Cpk = 0.67

c)

Yes it meets all specifications

Step-by-step explanation:

The specification for a plastic handle calls for a length of 6.0 inches ± .2 inches. The standard deviation of the process is estimated to be 0.05 inches. What are the upper and lower specification limits for this product? The process is known to operate at a mean thickness of 6.1 inches. What is the Cp and Cpk for this process?   Is this process capable of producing the desired part?

Given that:

Mean (μ) = 6.1 inches, Standard deviation (σ) = 0.05 inches and the length of the plastic handle is 6.0 inches ± .2

a) Since the length of the plastic handle is 6.0 inches ± .2  = (6 - 0.2, 6 + 0.2)

The Upper specification limits (USL) = 6 inches + 0.2 inches = 6.2 inches

The lower specification limits (LSL) = 6 inches - 0.2 inches = 5.8 inches

b) The Cp is given by the formula:

[tex]Cp=\frac{(USL-LSL)}{6\sigma} =\frac{(6.2-5.8)}{6*0.05} =1.33[/tex]

The Cpk is given by the formula:

c)

The upper specification limit lies about 3 standard deviations from the centerline, and the lower specification limit is further away, so practically all units will meet specifications

[tex]Cpk=min(\frac{USL-\mu}{3\sigma},\frac{\mu -LSL}{3\sigma})=min(\frac{6.2-6.1}{3*0.05},\frac{6.1-5.8}{3*0.05})=min(0.67,2)=0.67[/tex]

Final answer:

The upper specification limit for the plastic handle is 6.2 inches, and the lower specification limit is 5.8 inches, with these limits defining the acceptable range for the handle length.

Explanation:

The specification for a plastic handle is given as a length of 6.0 inches with a tolerance of ± 0.2 inches. This means the upper specification limit (USL) and the lower specification limit (LSL) are defined by adding and subtracting the tolerance to the target length respectively. The process standard deviation is 0.05 inches, but this does not affect the USL and LSL directly; it's a measure of the process variation.

The USL and LSL are calculated as follows:

These limits are the range within which the plastic handle lengths should fall according to the given specifications.

Without the provided number line, we cannot determine the exact relationship between f and g. Inequalities f < g or f > g represent f being less than or greater than g, respectively. To write the correct inequality, one must refer to the positions of f and g on the number line.

Since the number line is not provided, we cannot see the exact positions of f and g. However, we can discuss how to write inequalities to compare two integers based on their positions on a number line. If f is located to the left of g on the number line, it means that f is less than g. The inequality for this scenario would be f < g. On the other hand, if f is located to the right of g, then f is greater than g, and the corresponding inequality would be f > g.

You can use an inequality symbol to show how two metric measurements are related. If two numbers are the same, the inequality symbol would be the equal sign, representing they are equivalent. However, without the number line, we cannot determine the exact relationship between f and g, so one must look at the number line to ascertain the correct inequality to use.

Answer:

Josephine is 6 years old.

Step-by-step explanation:

6 * 5 = 30

12 * 3=  36

Josephine is currently 6 years old.

We are given two conditions about the ages of Josephine and her father.

Josephine’s father is 5 times as old as Josephine currently.

In 6 years, he will be only three times as old as she will be at that time.

Let's let 'J' represent Josephine's current age.

Then her father's current age would be 5J. In 6 years, Josephine will be J+6 and her father will be 5J+6.

According to the second condition, at that time her father's age will be three times Josephine's age.

So, we set up the equation 5J+6 = 3(J+6).

To find Josephine's current age, we solve the equation:

5J + 6 = 3J + 18

Subtract 3J from both sides:

2J + 6 = 18

Now subtract 6 from both sides:

2J = 12

And divide both sides by 2:

J = 6

Therefore, Josephine is currently 6 years old.

Answer: The answer is 1. Substitute (2,1) into x+3y=5

Step-by-step explanation: Remember, (2,1) 2 is substituted for x and 1 is used as the y substitute

Answer:

1. 2+3(1)=5

2. 5=5 is true

3. 1=-(2)+3

4. 1=1 is true

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 better explanation :)

Hope it helps!!!!!!!!!

Answer: 152

Step-by-step explanation:

305 - 153 = 152

The height of the Statue of Liberty itself, without the pedestal, is 152 feet. This is found by subtracting the pedestal height (153 feet) from the total height (305 feet).

The student is asked to determine the height of the Statue of Liberty excluding its pedestal. Given that the total height of the Statue of Liberty including its pedestal is labeled as 305 feet, and the total height is 153 feet more than the height of the statue alone, we can set up the following equation to solve for the height of the statue (let's call it S):

S + 153 = 305

To find the height of the Statue of Liberty without the pedestal, we subtract 153 from both sides of the equation:
S = 305 - 153
S = 152

Therefore, the height of the Statue of Liberty itself is 152 feet.

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Answer:

The correct answer is (E).

This is not an experiment because no treatment is being imposed upon the customers. Additionally, this study used a stratified sample because independent random samples were selected from two distinct populations of customers.

Step-by-step explanation:

The correct answer is (E) An observational study using a stratified sample.

Stratified sampling is also known as stratified random sampling. The stratified sampling process starts with researchers dividing a diverse population into relatively homogeneous groups called strata, the plural of stratum. Then, they draw a random sample from each group (stratum) and combine them to form their complete representative sample.

Given that a data of a survey, the study found that the mean satisfaction rating was significantly greater among customers that purchased computers that offer tech support.

We need to find which is the best description of this study,

This study used a stratified sample because independent random samples were selected from two distinct populations of customers.

This is not an experiment because no treatment is being imposed upon the customers.

Hence, the best description is an observational study using a stratified sample.

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The complete question is attached.

Answer: 35.5cm^2

Step-by-step explanation:

I believe we're talking about a rectangular. So it would be length * width or 7.5cm * 5cm = 35.5cm^2

Answer:

b) 0.9313

c) mean = 2.4

standard deviation= 1.3856

d) 0.5583

Step-by-step explanation:

Given:

p = 0.2

n = 12

a) X= number of applicants classified as deceptive.

Probability mass function of X will be:

[tex] P(X=x) = \left(\begin{array}{c}12\\x\end{array}\left) (0.2)^x(1-0.2)^1^2^-^x,x=0, 1, 2, .....,12 [/tex]

b) Probability that the polygraph says at least 1 is deceptive:

[tex] P(X≥1) = 1 -P(X=0) = 1 -\left(\begin{array}{c}12\\0\end{array}\left) (0.2)^0(1-0.2)^1^2^-^0[/tex]

= 1 - 0.0687

= 0.9313

c) The mean number among 12 truthful persons who will be classified as deceptive:

E(X) = n•p

= 12 * 0.2

= 24

Standard deviation:

[tex]s.d = \sqrt{12*0.2*(1-0.2)}[/tex]

= 1.3856

d) Probability that the number classified deceptive is less than the mean:

[tex] P(X<2.4) = P(X≤2) = E^2_x_=_0 \left(\begin{array}{c}12\\0\end{array}\left) (0.2)^0(1-0.2)^1^2^-^0 [/tex]

= 0.5583

Answer:

The number of Television at the beginning was 90

The number of video cassette recorders at the beginning was 89

Step-by-step explanation:

Selling Price of 1 Television =$360.

Selling Price of 1 video cassette recorders for $270.

Let the number of Television at the beginning=x

Let the number of video cassette recorders at the beginning =y

Opening Stock =$56,430.

Therefore:

It sells three quarters of the televisions and one third of the video cassette recorders for a total of $32,310.

[tex]\frac{3}{4}(360)x+\frac{1}{3}(270)y= \$32,310[/tex]

270x+90y=32310

We then solve the two equations to obtain x and y.

97200x+72900y=15236100

97200x+32400y=11631600

Subtract

40500y=3604500

y=89

Substitute y=89 into 270x+90y=32310 to obtain x

270x+90(89)=32310

270x=32310-8010=24300

x=90

Therefore:

The number of Television at the beginning was 90

The number of video cassette recorders at the beginning was 89

Answer:

90 televisions and 89 video cassette recorders

Step-by-step explanation:

The unit cost of television = $360

The unit cost of video cassette recorder = $270

Let "T" represent the number of televisions and "R" represent the number of recorders, so that we can make representations using equations from the statements.

At the beginning of the week, Total Stock is worth $56,430, where

Total Stock = Total cost of televisions + Total cost of recorders

Total Cost = Unit Cost × Number of items

$56,430 = 360T + 270R  This is the first equation

Next, During the week, Number of Sales = [tex]\frac{3}{4}[/tex] T + [tex]\frac{1}{3}[/tex] R

Total Sales Price = 360 ([tex]\frac{3}{4}[/tex] T) + 270 ([tex]\frac{1}{3}[/tex] R)

$32,310 = 270T + 90R     This is the second equation

Solving both equations simultaneously, let us use elimination method which involves equating one of the two terms in both equations. Let us multiply the second equation by 3. This doesn't affect the equation, since we are doing it to all the terms in it.

56,430 = 360T + 270R

32,310 = 270T + 90R            × 3

So, we have;

56,430 = 360T + 270R

96,930 = 810T + 270R

Subtracting both equations, we have;

96,930 - 56,430 = 810T - 360T

40,500 = 450T

T = [tex]\frac{40,500}{450}[/tex] = 90

Since we now have the number of televisions, we can get the number of recorders by putting 90 in any (say, the second) equation.

32,310 = 270 (90) + 90R

32,310 = 24,300 + 90R

32,310 - 24,300 = 90R

8010 = 90R

R = [tex]\frac{8010}{90}[/tex] = 89

At the beginning of the week, the store had 90 televisions and 89 video cassette recorders

Answer:

[tex] P(1.28< X< 3.8) [/tex]

And we can use the z score formula to calculate how many deviations we are within the mean

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

And if we use this formula we got:

[tex] z = \frac{1.28-2.54}{0.42}= -3[/tex]

[tex] z = \frac{3.8-2.54}{0.42}= 3[/tex]

And using the empirical rule we know that within 3 deviation from the mean we have 99.7% of the values

Step-by-step explanation:

Previous concepts

The empirical rule, also known as three-sigma rule or 68-95-99.7 rule, "is a statistical rule which states that for a normal distribution, almost all data falls within three standard deviations (denoted by σ) of the mean (denoted by µ)".

Let X the random variable who represent the grade point averages of undergraduate students.

From the problem we have the mean and the standard deviation for the random variable X. [tex]E(X)=2.54, Sd(X)=0.42[/tex]

So we can assume [tex]\mu=2.54 , \sigma=0.42[/tex]

On this case in order to check if the random variable X follows a normal distribution we can use the empirical rule that states the following:

• The probability of obtain values within one deviation from the mean is 0.68

• The probability of obtain values within two deviation's from the mean is 0.95

• The probability of obtain values within three deviation's from the mean is 0.997

For this case we want to find this probability:

[tex] P(1.28< X< 3.8) [/tex]

And we can use the z score formula to calculate how many deviations we are within the mean

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

And if we use this formula we got:

[tex] z = \frac{1.28-2.54}{0.42}= -3[/tex]

[tex] z = \frac{3.8-2.54}{0.42}= 3[/tex]

And using the empirical rule we know that within 3 deviation from the mean we have 99.7% of the values

Approximately 99.7 percent of the students have grade point averages between 1.28 and 3.8 according to the empirical rule.

The empirical rule states that approximately 68 percent of the data lies within one standard deviation of the mean, 95 percent lies within two standard deviations, and more than 99 percent lies within three standard deviations. In this case, the mean grade point average is 2.54 and the standard deviation is 0.42. To find the percentage of students with grade point averages between 1.28 and 3.8, we need to find the z-scores for both values and calculate the area under the curve between those z-scores.

First, we find the z-score for 1.28 using the formula: z = (x - µ) / σ = (1.28 - 2.54) / 0.42 = -3.0476. Then, we find the z-score for 3.8 using the same formula: z = (3.8 - 2.54) / 0.42 = 3.0476. Using a standard normal distribution table or a calculator, we can find the area between these two z-scores, which is approximately 99.7 percent.

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Answer:

so the common multiple of 4 and 6 is 2  i hope this helps!   :)

Step-by-step explanation:

well the common multiples of 4 and 6 is 2

2 * 2 = 4

2 * 3 = 6

so there has to be a certain number that goes into both of the numbers you are working with

Answer:

Common multiples of 4 and 6 are 12, 24, 36, 48, 60, 72, 84, 96...

Answer: a) 1.029, b) (-5.318, -1.282).

Step-by-step explanation:

Since we have given that

[tex]n_1=13\\\\n_2=17\\\\\bar{x_1}=1.9\\\\\bar{x_2}=5.2[/tex]

and

[tex]s_1=2.1\\\\s_2=3.5[/tex]

So, the standard error for comparing the means :

[tex]SE=\sqrt{\dfrac{s^2_1}{n_1}+\dfrac{s^2_2}{n_2}}\\\\SE=\sqrt{\dfrac{2.1^2}{13}+\dfrac{3.5^2}{17}}\\\\SE=\sqrt{1.0598}\\\\SE=1.029[/tex]

At 95% confidence interval, z = 1.96

So, Confidence interval would be

[tex]\bar{x_1}-\bar{x_2}\pm z\times SE\\\\=(1.9-5.2)\pm 1.96\times 1.0294\\\\=-3.3\pm 2.017624\\\\=(-3.3-2.018,-3.3+2.018)\\\\=(-5.318,-1.282)[/tex]

Hence, a) 1.029, b) (-5.318, -1.282).

Answer:

a) The probability that exactly 41 of the murders were cleared is 0.0013

b) The probability that between 36 and 38 of the murders, inclusive, were cleared is 0.0809.

c) Yes, it would be unusual

Step-by-step explanation:

Let p=62% considered as the probability of having a commited that is cleared by arres or exceptional means. We assume that choosing each of the 50 commited is independent of each choose. Then, let X be the number of cleared. In this case, X is distributed as a binomial random variable. Recall that, in this case,

[tex] P(X=k) = \binom{50}{k} p^{k}(1-p)^{50-k}[/tex] for[tex]0\leq k \leq 50[/tex], with p=0.62

a) We have that

[tex] P(X=40) = \binom{50}{40} p^{40}(1-p)^{50-40} =0.001273487  [/tex]

b) We are asked for the following

[tex]P(36\leq X \leq 38) = P(X=36)+P(X=37)+P(X=38) = 0.080888936

[/tex] (The specific calculation is omitted.

c) We will check for the following probability [tex]P(X\leq 19)[/tex]

[tex]P(X\leq 19 ) = \sum_{k=0}^{19} P(X=k) = 0.000499222 [/tex]

Given that the probability of this event is really close to 0, it would be unusual if less than 19 murders are cleared.

The question deals with the application of binomial probability distribution in a real life situation involving crime investigation. The probability values for a certain range or exact number of cleared murders can be calculated by using the binomial probability formula. It would be statistically unusual for fewer than 19 murders to be cleared given a 62% clearance rate.

This question can be approached using the binomial distribution, where the number of successes in a sequence of n independent experiments (in this case, the number of murders being cleared) follows a binomial distribution.

a. The probability that exactly 41 of the murders were cleared can be found by calculating the binomial probability. This can be done by using the formula: P(X=k) = C(n, k) * (p^k) * ((1-p)^(n-k)). In this case, n=50 (number of trials/murders), k=41 (number of successes/murders cleared), and p=0.62 (probability of success/clearing a murder). You need to substitute these values into the formula and calculate the value.

b. Finding the probability that between 36 and 38 murders were cleared involves the same process, but you need to calculate for k=36, 37, and 38, and then add the results together to get the total probability.

c. If fewer than 19 of the murders were cleared, it would be statistically unusual considering the 62% clearance rate according to the survey. The reasoning being that, given a 62% probability, the expectation would be significantly higher.

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Answer:

And the F statistic calculated from the mean squares if [tex]F_{calc}[/tex]. And for this case we know that [tex] F_{calc}>F_{critical}[/tex]. So then we can reject the null hypothesis that all the means are equal at a significance level given [tex]\alpha[/tex]. And the best conclusion would be:

reject H0 since there is evidence all the means differ.

Step-by-step explanation:

Previous concepts

Analysis of variance (ANOVA) "is used to analyze the differences among group means in a sample".  

The sum of squares "is the sum of the square of variation, where variation is defined as the spread between each individual value and the grand mean"  

Solution to the problem

The hypothesis for this case are:

Null hypothesis: [tex]\mu_{A}=\mu_{B}=....=\mu_{k}[/tex]

Alternative hypothesis: Not all the means are equal [tex]\mu_{i}\neq \mu_{j}, i,j=A,B,...,k[/tex]

If we assume that we have [tex]p[/tex] groups and on each group from [tex]j=1,\dots,p[/tex] we have [tex]n_j[/tex] individuals on each group we can define the following formulas of variation:  

[tex]SS_{total}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x)^2 [/tex]  

[tex]SS_{between}=SS_{model}=\sum_{j=1}^p n_j (\bar x_{j}-\bar x)^2 [/tex]  

[tex]SS_{within}=SS_{error}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x_j)^2 [/tex]  

And we have this property  

[tex]SST=SS_{between}+SS_{within}[/tex]  

And the F statistic calculated from the mean squares if [tex]F_{calc}[/tex]. And for this case we know that [tex] F_{calc}>F_{critical}[/tex]. So then we can reject the null hypothesis that all the means are equal at a significance level given [tex]\alpha[/tex]. And the best conclusion would be:

reject H0 since there is evidence all the means differ.

In a one-way ANOVA, if the computed F statistic is greater than the critical F value you may eject H0 since there is evidence all the means differ.

F test is a method used in statistics to determine which models best fits the population from which the sample is derived.

The formula for calculating  the F-test statistic = explained variance / unexplained variance

The null hypothesis usually states that the means are the means are the same while the alternative hypothesis states that the means are not the same.

To learn more about the null hypothesis, please check: brainly.com/question/4454077

Answer:

17.38m

Step-by-step explanation:

Answer:

17.38m

Step-by-step explanation:

Answer:

52 weeks

365 days

10 years

100 years

Step-by-step explanation:

Answer:

1/4; 25%

Step-by-step explanation:

Both events happen with probability 1/2: there are 3 even numbers and 3 odd numbers in a die.

Since the two events are also independent (the outcome of the first die doesn't affect the outcome of the second), we have to multiply those probability.

So, you roll an odd number on the first die and an even number on the second die with probability

[tex]\dfrac{1}{2}\cdot\dfrac{1}{2}=\dfrac{1}{4}[/tex]

Answer:

The range of the p-value is: 0.050 < p-value < 0.100.

Step-by-step explanation:

For checking the equivalence of two population variances of independent samples, we use the f-test.

The test statistic is given by:

[tex]F=\frac{S_{1}^{2}}{S_{2}^{2}}\sim F_{\alpha, (n_{1}-1)(n_{2}-1)}[/tex]

It is provided that the hypothesis test is one-tailed.

The computed value of the test statistic is:

F = 4.23.

The degrees of freedom of the numerator and denominator are:

[tex]df_{1}=4\\df_{2}=5[/tex]

Use MS-Excel to compute the p-value as follows:

Step 1: Select function fX → F.DIST.RT.

Step 2: A dialog box will open. Enter the values of f-statistic and the two degrees of freedom.

*See the attachment below.

Step 3: Press OK.

The p-value is, 0.0728.

The range of the p-value is:

0.050 < p-value < 0.100

Answer:EX = RT ln [X]o

.........zF.....[X]i

EX = (1.987 cal/deg.mol)(293 deg) ln [X]o

.........z(23,062 cal/volt.mol)................[X]i

OR

EX = (8.315 joules/deg.mol)(293 deg) ln [X]o

.........z(96,485 joules/volt.mol)................[X]i

EK+ = 0.025 ln(12/400) = -0.088 V = -88 mV

ENa+ = 0.025 ln(450/55) = 0.053 V = 53 mV

ECa+2 = 0.0126 ln(10/0.0001) = 0.145 V = 145 mV

ECl- = -0.025 ln(550/56) = -0.058 V = -58 mV

Step-by-step explanation:

The Goldman equation is used to calculate resting membrane potential (Vm) considering the relative permeabilities of different ions. Given the permeabilities, the resting Vm can be estimated. If the K+ permeability increases, Vm will move closer to the Potassium equilibrium potential (EK).

The resting membrane potential (Vm) can be calculated using the Goldman equation, which considers the relative permeabilities and concentrations of different ions. The equation is: Vm = 61.5 log ((PK[K+]out + PNa[Na+]out + PCl[Cl-]in) / (PK[K+]in + PNa[Na+]in + PCl[Cl-]out)), where Px indicates the relative permeability of each ion & the square brackets contain the ion concentrations inside (in) or outside (out) the cell.

Given the permeabilities PK : PNa : PCl = 1.0 : 0.04 : 0.45, assuming concentrations inside and outside the cell in a balanced condition with no net movement of any ion, you might estimate the resting Vm.  

If the K+ permeability increases, Vm would reportedly move closer towards the Potassium equilibrium potential (EK). This is because the membrane is becoming more permeable to K+ and less responsive to the influences of other ions.

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