If the average score on this test is 510 with a standard deviation of 100 points, what percentage of students scored below 300? Enter as a percentage to the nearest tenth of a percent.

Answer:

Probability that a randomly selected student scored less than 156 on the final exam is 0.97725 .

Step-by-step explanation:

We are given that the exam scores on a statistics final exam are normally distributed with a mean of 140 points out of 200 and standard deviation of 8.

Let X = Score of students in exam

So, X ~ N([tex]\mu = 140, \sigma^{2} =8^{2}[/tex])

The z score probability distribution is given by;

           Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ N(0,1)

where, [tex]\mu[/tex] = population mean = 140

            [tex]\sigma[/tex] = standard deviation = 8

So, the probability that a randomly selected student scored less than 156 on the final exam is given by = P(X < 156)

 P(X < 156) = P( [tex]\frac{X-\mu}{\sigma}[/tex] < [tex]\frac{156-140}{8}[/tex] ) = P(Z < 2) = 0.97725 .

Therefore, the probability that a randomly selected student scored less than 156 on the final exam is 0.97725 .

To find the probability, calculate the z-score and look it up in a table or use a calculator. The probability of scoring less than 156 is approximately 0.9772.

To find the probability that a randomly selected student scored less than 156 on the final exam, we will use the z-score formula. The z-score formula is calculated by subtracting the mean from the given score and dividing by the standard deviation. In this case, the z-score is (156 - 140) / 8 = 2.

To find the probability, we can look up the z-score in a standard normal distribution table or use a calculator like the TI-84. Using either method, we find that the probability of a student scoring less than 156 is approximately 0.9772.

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Answer:

[tex]X \sim N(100,15)[/tex]  

Where [tex]\mu=100[/tex] and [tex]\sigma=15[/tex]

We select a sample of n=16 and we are interested on the distribution of [tex]\bar X[/tex], since the distribution for X is normal then we can conclude that the distribution for [tex] \bar X [/tex] is also normal and given by:

[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]

Because by definition:

[tex] \bar X = \frac{\sum_{i=1}^n X_i}{n}[/tex]

[tex] E(\bar X) = \mu[/tex]

[tex] Var(\bar X) = \frac{\sigma^2}{n}[/tex]

And for this case we have this:

[tex] \mu_{\bar X}= \mu = 100[/tex]

[tex] \sigma_{\bar X} = \frac{15}{\sqrt{16}}= 3.75[/tex]

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 IQ scores of a population, and for this case we know the distribution for X is given by:

[tex]X \sim N(100,15)[/tex]  

Where [tex]\mu=100[/tex] and [tex]\sigma=15[/tex]

We select a sample of n=16 and we are interested on the distribution of [tex]\bar X[/tex], since the distribution for X is normal then we can conclude that the distribution for [tex] \bar X [/tex] is also normal and given by:

[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]

Because by definition:

[tex] \bar X = \frac{\sum_{i=1}^n X_i}{n}[/tex]

[tex] E(\bar X) = \mu[/tex]

[tex] Var(\bar X) = \frac{\sigma^2}{n}[/tex]

And for this case we have this:

[tex] \mu_{\bar X}= \mu = 100[/tex]

[tex] \sigma_{\bar X} = \frac{15}{\sqrt{16}}= 3.75[/tex]

Answer:

198 students and 127 adults

Step-by-step explanation:

Answer:

a) 0.931491

b) 0.995307

c) 0.994030

Step-by-step explanation:

a) Since all components must be working, the reliability of the computer is the product of the reliability of the three components:

[tex]R_1 = 0.97*0.97*0.99\\R_1=0.931491[/tex]

b) The resulting reliability is now the reliability of the first computer, added to the possibility of failure of the first computer multiplied by the reliability of the second computer:

[tex]R= R_1 +(1-R_1)*R_2\\R= 0.931491+(1-0.931491)*0.97*0.97*0.99\\R=0.995307[/tex]

c) If a switch with reliability of 0.98 must be activated to turn on the second computer, the switch's reliability must be taken into account as follows:

[tex]R= R_1 +(1-R_1)*R_2*R_S\\R= 0.931491+(1-0.931491)*0.97*0.97*0.99*0.98\\R=0.994030[/tex]

The reliability of the system is simply its probability of not failing

(a) The reliability of the computer

This is the product of the reliabilities of the three modules.

So, we have:

[tex]R = 0.97 \times 0.97 \times 0.99[/tex]

[tex]R = 0.931491[/tex]

Approximate

[tex]R = 0.9315[/tex]

Hence, the reliability of the computer is 0.9315

(b) The reliability when a backup is used

In (a), the reliability of the computer is 0.9315

When the computer fails, the reliabilities of the other two are 1 - 0.9315 and 1 - 0.9315.

So, the reliability when a backup is used is calculated using the following complement rule

[tex]R = 1 - [(1 - 0.9315) \times (1 - 0.9315)][/tex]

[tex]R = 0.99530775[/tex]

Approximate

[tex]R = 0.9953[/tex]

Hence, the reliability of the computer when a backup is 0.9953

(c) The overall reliability of the system

In (a), the reliability of the computer is 0.9315.

Also, the reliability of the switch is 0.98

So, the reliability of the backup is:

[tex]R = 0.9315 \times 0.98[/tex]

[tex]R = 0.9129[/tex]

So, the overall system has:

The reliability of the overall system is then calculated using the following complement rule

[tex]R = 1 - [(1 - 0.9315) \times (1 - 0.9129)][/tex]

[tex]R = 0.9940[/tex]

Hence, the reliability of the overall system is 0.9940

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Answer:

I code an example question with answer.

Step-by-step explanation:

A study on educational aspirations of High School Students ( S. Crysdale, International Journal of Comparative Sociology, Vol 16, 1975, pp 19-36) measured aspirations using the scale (some high school, high school graduate, some college, college graduate). For students whose family income was low, the counts in these categories were (9, 44, 13, 10); when family income was middle, the counts were (11, 52, 23, 22); when family income was high, the counts were (9, 41, 12, 27)

A. Construct a suitable contingency table for the above data.

B. Find the conditional distribution on aspirations for those whose family income was high.

C. Conduct a Chi-square test of Independence between educational aspirations and income levels.

D. Explain what further analyses you could do that would be more informative than a chi-squared test.

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Chi-Squared Test for Homogeneity of Several Categorical Populations

Null Hypothesis: populations of people are homogeneous with respect to the four levels of education (low, med, high income groups educated same)

Alternative Hypothesis: populations not homogeneous.

Chi-Square Test: Some High School, Grad High School, Some College, Grad College

Expected counts are printed below observed counts

Chi-Square contributions are printed below expected counts

Chi-Square Test: Some High School, Grad High School, Some College, Grad College

Expected counts are printed below observed counts

Chi-Square contributions are printed below expected counts

Some Grad

High High Some Grad

School School College College Total

Low Income

9 44 13 10 76 [observed counts]

8.07 38.14 13.36 16.42 [expected counts]

0.106 0.901 0.010 2.513 [Chi-Square contr]

Medium Income

11 52 23 22 108

11.47 54.20 18.99 23.34

0.019 0.089 0.847 0.077

High Income

9 41 12 27 89

9.45 44.66 15.65 19.23

0.022 0.300 0.851 3.135

Total 29 137 48 59 273

Chi-Sq = 8.871, DF = 6, P-Value = 0.181

"P-Value" = 0.181 > 0.10 [90% confidence interval]; thus Null Hypothesis of homogeneity should be rejected (low, med, high income groups educated same). Alternative Hypothesis should be accepted (education dependent upon income level)

The task is a statistical analysis using SAS/R to test the independence of two variables, economic status and educational aspirations. This involves chi-square and likelihood ratio tests, investigating standardized residuals, and performing more powerful tests for comprehensive results.

The question involves a statistical analysis task using SAS/R to determine the independence between two variables: economic status and educational aspirations. The Chi-square (X2) and likelihood ratio (G2) tests can be utilized to analyze the independence. Standardized residuals can help diagnose potential patterns and significance of divisions, while more powerful tests such as Fisher's exact test or Monte Carlo simulation could provide further insights.

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Answer:

0.15651

Step-by-step explanation:

This can be approximated using a Poisson distribution formula.

The Poisson distribution formula is given by

P(X = x) = (e^-λ)(λˣ)/x!

P(X ≤ x) = Σ (e^-λ)(λˣ)/x! (Summation From 0 to x)

where λ = mean of distribution = 20 red bags of skittles (20% of 100 bags of skittles means 20 red bags of skittles)

x = variable whose probability is required = less than 16 red bags of skittles

P(X < x) = Σ (e^-λ)(λˣ)/x! (Summation From 0 to (x-1))

P(X < 16) = Σ (e^-λ)(λˣ)/x! (Summation From x=0 to x=15)

P(X < 16) = P(X=0) + P(X=1) + P(X=2) +......+ P(X=15)

Solving this,

P(X < 16) = 0.15651

The question asks for the probability that the proportion of red Skittles is less than 16% from a sample of 100 bags. This involves the application of sampling distribution of a sample proportion, wherein the distribution should follow a normal distribution if certain conditions are met. The question can be solved by calculating a z-score and finding the corresponding probability from a standard normal distribution table.

This question involves the concept of sampling distribution of a sample proportion. Here, under certain conditions, the sampling distribution of p' (the sample proportion) tends to follow a normal distribution. The mean (expected value) of the distribution is equal to the population proportion (p), and the standard deviation (standard error) of the distribution is sqrt [ p(1 - p) / n ], where n is the size of the sample.

Given that the population proportion (p) = 0.20 and n = 100. We are asked to find the probability that the sample proportion (p') is less than 0.16 (16%). So, we can represent this situation as: P( p' < 0.16 ).

To find this probability we need to standardize our value of interest (0.16), resulting in a z-score. The z-score = ( p' - p ) / sqrt [ p(1 - p) / n ]. Plugging our values in, you can calculate the z-score, and apply it into a standard normal distribution table or use a calculator that can calculate probabilities using normal distribution to find the probability.

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The growth factor for the mass of bacteria is 2.8 in 1 week.

Step-by-step explanation:

Step 1:

It is given that the bacteria in the culture increase by 260% per week. Now we assign values to get a better understanding. If there were 1,000 bacteria in the culture at the start of the week there would be 1,000 + 260% at the week's end = 1,000 + 2,600 = 3,600 at the week's end.

Step 2:

To calculate the growth factor, we calculate the difference between the present value and the past value and divide it by the past value.

Here the past value is 1,000 and the present value is 2,600.

Growth factor = [tex]\frac{difference in values}{past value} = \frac{3,600-1,000}{1000} = \frac{2,600}{1,000} = 2.6.[/tex]

So the growth factor is 2.6 per week.

Answer:

6.08333

Step-by-step explanation:

6ft 0.8inches

Answer:

6 ft 1 in

Step-by-step explanation:

Answer:

P

Step-by-step explanation:

(all 5 correctly) = (5 chosen among 7) / (5 chosen among 10)

Answer:

The point estimate used to estimate the mean height of all adult males in Idaho was 69.505 inches.

Step-by-step explanation:

The point estimate is the halfway point of the confidence interval, that is, the lower bound added to the upper bound, and then this sum is divided by 2. So

Lower bound: 62.535

Upper bound: 76.478

Point estimate:

[tex]P_{e} = \frac{62.535 + 76.478}{2} = 69.505[/tex]

The point estimate used to estimate the mean height of all adult males in Idaho was 69.505 inches.

The point estimate used to estimate the mean height of all adult males in Idaho was 69.505 inches.

Since Given that a 90% confidence interval for the mean height of all adult males in Idaho measured in inches was [62.532, 76.478].

So, the estimation of the point is

[tex]= (62.532 + 76.478) \div 2[/tex]

= 69.505 inches

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Answer:

Step-by-step explanation:

Hello!

The variable of interest is X: number of individuals that hold multiple jobs in a sample of 225.

And you need to calculate the probability of the proportion of individuals being between 0.14 and 0.18.

Considering that the parameter of interest is the proportion and the sample is large enough you can use the standard normal approximation to calculate this interval.

[tex]Z= \frac{p'-p}{\sqrt{\frac{p(1-p)}{n} } }[/tex]

Unfortunately, there is no given value for the population proportion of individuals that have multiple jobs. Let's say, for example, that his proportion is 10%.

You can symbolize the probabilities as:

P(0.14≤X≤0.18)= P(X≤0.18)-P(X≤0.14)

Using the approximation of the standard normal you can standardize these proportion values:

P(Z≤(0.18-0.1)/√[(0.1*0.9)/225])-P(Z≤(0.14-0.1)/√[(0.1*0.9)/225])

P(Z≤4)-P(Z≤2)

Now you have to look for the corresponding values in the table of the Z distribution (right table, positive numbers of Z)

P(Z≤4)-P(Z≤2)= 1 - 0.97725= 0.02275

I hope it helps!

Answer:

0.1276

Step-by-step explanation:

Let sample space = S

∴ n(S) = 225

Probability of multiple jobs A, that is, P(A) = 0.14

Probability of multiple jobs B, that is, P(B) = 0.18

Hence, P(A) = n(A)/n(S)

∴ 0.14 = n(A)/225

n(A) = 0.14 X 225 = 31.5%

Also, P(B) = n(B)/n(S)

∴ 0.18 = n(B)/225

n(B) = 0.18 X 225 = 40.5%

The probability that the multiple jobs, P( A2 ∩ B2) = P(A2) X P(B2)

P(A2) = 0.315 and P(B2) = 0.405

∴ P(A2 ∩ B2) =  P(A2) X P(B2) = 0.315 X 0.405 = 0.1276

Elaborating on the matching of the sampling method descriptions:

To summarize, the correct matching would be:

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1: d. Convenience

2: e. Simple random

3: c. Cluster

4: a. Stratified

5: b. Systematic

1. Convenience sampling (d):

2. Simple random sampling (e):

3. Cluster sampling (c):

4. Stratified sample(s):

5. Systematic sampling (b):

Complete question:

Match the names of the given sampling method descriptions.

Situation:

1. drawing 50 names from a hat

2. ask all the students in your math class

3. randomly select two tables in the dining room and examine all the people at those two tables

4. dividing all students according to grades and selecting 10 students from each grade

5. call every 15th phone number on every 5th page of the phone book

Sampling method:

a. Stratified

b. Systematic

c. Cluster

d. Convenience

e. Simple random

Answer:

[tex]z=\frac{0.458 -0.35}{\sqrt{\frac{0.35(1-0.35)}{83}}}=2.06[/tex]  

[tex]p_v =P(z>2.063)=0.020[/tex]  

So the p value obtained was a low value and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of students with jobs is significantly higher than 0.35.

Step-by-step explanation:

Data given and notation

n=83 represent the random sample taken

X=38 represent the students with jobs

[tex]\hat p=\frac{38}{83}=0.458[/tex] estimated proportion of students with jobs

[tex]p_o=0.35[/tex] is the value that we want to test

[tex]\alpha=0.05[/tex] represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

[tex]p_v[/tex] represent the p value (variable of interest)  

Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the true proportion is higher than 0.35:  

Null hypothesis:[tex]p \leq 0.35[/tex]  

Alternative hypothesis:[tex]p > 0.35[/tex]  

When we conduct a proportion test we need to use the z statistic, and the is given by:  

[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)  

The One-Sample Proportion Test is used to assess whether a population proportion [tex]\hat p[/tex] is significantly different from a hypothesized value [tex]p_o[/tex].

Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

[tex]z=\frac{0.458 -0.35}{\sqrt{\frac{0.35(1-0.35)}{83}}}=2.06[/tex]  

Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The significance level provided [tex]\alpha=0.05[/tex]. The next step would be calculate the p value for this test.  

Since is a right tailed test the p value would be:  

[tex]p_v =P(z>2.06)=0.020[/tex]  

So the p value obtained was a low value and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of students with jobs is significantly higher than 0.35.

The student's question pertains to calculating the regression line equation that represents the relationship between rainfall and wheat yield. This can be done using a calculator's statistical functions to provide the least-squares regression line equation, which allows for the prediction of one variable based on another.

The question asks about finding the equation for a linear regression line based on the relationship between rainfall and wheat yield. To accomplish this, one must enter the given data into a calculator, generate a scatter plot, and then use the calculator's regression function to find the least-squares regression line's equation. This regression equation typically takes the form î = a + bx, where 'a' is the y-intercept and 'b' is the slope of the line.

In terms of finding the regression line manually, a commonly used method is the median-median line approach, but for this question, we are instructed to use a calculator. The process involves statistical analysis to determine the line that minimally deviates from all data points in the scatter plot. Once the equation is calculated, one can predict values such as wheat yield for a given amount of rainfall.

a) Probability that all four have type B+ blood = 0.00031213

b) Probability that none of the four have type B+ blood = 0.57289761

c) Probability that at least one of the four has type B+ blood = 0.42710239

d) B. The event in part (a) is unusual because its probability is less than or equal to 0.05.

To solve these probability problems, we'll use the binomial probability formula:

[tex]P(X=k) = (n, k) \times p^k \times (1-p)^{(n-k)[/tex]

Where:

P(X=k) is the probability of having exactly k successes in n trials.

(n choose k) is the number of ways to choose k successes from n trials (n! / (k! (n-k)!), where n! is the factorial of n).

p is the probability of success (having type B+ blood in this case).

q = 1 - p is the probability of failure (not having type B+ blood).

n is the number of trials.

Given:

p = 0.13 (probability of having type B+ blood)

q = 1 - p = 0.87 (probability of not having type B+ blood)

n = 4 (number of trials)

Let's solve each part step by step:

(a) Probability that all four have type B+ blood:

[tex]P(X=4) = (4, 4) \times 0.13^4 \times 0.87^{(4-4)[/tex]

[tex]P(X=4) = 1 \times 0.00031213 \times 1 \\\\= 0.00031213[/tex]

(b) Probability that none of the four have type B+ blood:

[tex]P(X=0) = (4, 0) \times 0.13^0 \times 0.87^4 \\\\P(X=0) = 1 \times 1 \times 0.57289761 \\\\= 0.57289761[/tex]

(c) Probability that at least one of the four has type B+ blood:

P(at least one) = 1 - P(none)

P(at least one) = 1 - 0.57289761

= 0.42710239

Now, let's determine which events are considered unusual. Generally, an event with a probability less than or equal to 0.05 is considered unusual.

Let's compare the probabilities:

Probability in part (a): 0.00031213 (less than 0.05)

Probability in part (b): 0.57289761 (greater than 0.05)

Probability in part (c): 0.42710239 (greater than 0.05)

Based on the comparison, the only event that can be considered unusual is the event in part (a) because its probability is less than 0.05. Therefore, the correct answers are:

B. The event in part (a) is unusual because its probability is less than or equal to 0.05.

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The calculations show that the probability of all four people having B+ blood is 0.000028561, and the likelihood of none of them having B+ blood is 0.569532. The chance of at least one of them having B+ blood is 0.430468. Thus, only the event in part (a) is considered unusual due to its low probability.

This question is about using probability principles to figure out the likelihood of having certain blood types in a population.

(a) To find the probability that all four individuals have type B+ blood, we need to multiply the individual probabilities together. The probability that one person has B+ blood is given as 13% or 0.13. So, the probability that all four have B+ blood is 0.13*0.13*0.13*0.13 = 0.000028561.

(b) The probability that none of the four have type B+ blood is the complement of the probability that one person has B+ blood. This is 1 - 0.13 = 0.87. We now raise this to the power of four to find the probability that all four selected people do not have B+ blood: 0.87*0.87*0.87*0.87 = 0.569532.

(c) The probability that at least one has type B+ blood is the complement of the result in part b. We subtract our answer from part b from 1: 1 - 0.569532 = 0.430468.

(d) An event is considered unusual if its probability is less than or equal to 0.05. Here, the event in part (a) is unusual because its probability (0.000028561) is less than or equal to 0.05.

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Answer:

  decreasing, non-increasing

Step-by-step explanation:

The sequence is ...

  1, 1/2, 1/3, 1/4, ...

Each term is smaller than the one before it, so the sequence is decreasing (also, non-increasing).

The sum of the squared residuals of the least squares line for the given data is 8691.90.

To compute the sum of the squared residuals for the given data, we will perform the steps of calculating the least squares line manually. The least squares line represents the linear regression line that minimizes the sum of the squared differences between the observed data points and the predicted values.

Week 1: Number of Grunts = 90, Age (days) = 125

Week 2: Number of Grunts = 68, Age (days) = 141

Week 3: Number of Grunts = 39, Age (days) = 155

Week 4: Number of Grunts = 44, Age (days) = 160

Week 5: Number of Grunts = 63, Age (days) = 167

Week 6: Number of Grunts = 40, Age (days) = 174

Week 7: Number of Grunts = 62, Age (days) = 183

Week 8: Number of Grunts = 17, Age (days) = 189

Week 9: Number of Grunts = 20, Age (days) = 195

Calculate the means of the Number of Grunts (Y) and Age (X) variables:

Mean of Y (Number of Grunts) = (90 + 68 + 39 + 44 + 63 + 40 + 62 + 17 + 20) / 9 = 48.78

Mean of X (Age) = (125 + 141 + 155 + 160 + 167 + 174 + 183 + 189 + 195) / 9 = 165

Calculate the deviations from the means for each data point:

For each data point, subtract the mean of X (Age) from the specific X value and the mean of Y (Number of Grunts) from the specific Y value.

Deviation from mean for each X value:

125 - 165.67 = -40.67

141 - 165.67 = -24.67

155 - 165.67 = -10.67

160 - 165.67 = -5.67

167 - 165.67 = 1.33

174 - 165.67 = 8.33

183 - 165.67 = 17.33

189 - 165.67 = 23.33

195 - 165.67 = 29

Deviation from mean for each Y value:

90 - 48.78 = 41.22

68 - 48.78 = 19.22

39 - 48.78 = -9.78

44 - 48.78 = -4.78

63 - 48.78 = 14.22

40 - 48.78 = -8.78

62 - 48.78 = 13.22

17 - 48.78 = -31.78

20 - 48.78 = -28.78

Calculate the sum of the products of the deviations:

Sum of (Deviation from mean for X * Deviation from mean for Y)

= (-40.67 * 41.22) + (-24.67 * 19.22) + (-10.67 * -9.78) + (-5.67 * -4.78) + (1.33 * 14.22) + (8.33 * -8.78) + (17.33 * 13.22) + (23.33 * -31.78) + (29.33 * -28.78)

= -5.02 + -0.90 + 1.04 + 0.27 + 18.95 + -73.14 + 228.44 + -739.97 + -845.44

= -1407.77

Calculate the sum of the squared deviations for X:

Sum of (Deviation from mean for X)^2

= (-40.67)^2 + (-24.67)^2 + (-10.67)^2 + (-5.67)^2 + (1.33)^2 + (8.33)^2 + (17.33)^2 + (23.33)^2 + (29.33)^2

= 16572.86

Calculate the sum of the squared residuals:

Sum of squared residuals = Sum of (Deviation from mean for Y)^2

= (41.22)^2 + (19.22)^2 + (-9.78)^2 + (-4.78)^2 + (14.22)^2 + (-8.78)^2 + (13.22)^2 + (-31.78)^2 + (-28.78)^2

= 8691.90

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The sum of the squared residuals of the least squared line for the given data is 8399.

To compute the sum of the squared residuals of the least squared line by hand, you will need to follow these steps:

1. Start by organizing the given data into two columns: one for the number of grunts and one for the age of the warthog.

Week 1: Number of Grunts = 90, Age (days) = 125

Week 2: Number of Grunts = 68, Age (days) = 141

Week 3: Number of Grunts = 39, Age (days) = 155

Week 4: Number of Grunts = 44, Age (days) = 160

Week 5: Number of Grunts = 63, Age (days) = 167

Week 6: Number of Grunts = 40, Age (days) = 174

Week 7: Number of Grunts = 62, Age (days) = 183

Week 8: Number of Grunts = 17, Age (days) = 189

Week 9: Number of Grunts = 20, Age (days) = 195

2. Calculate the mean (average) of the age and the number of grunts. To do this, add up all the values in each column and divide by the total number of data points.

Mean of the age (days):

(125 + 141 + 155 + 160 + 167 + 174 + 183 + 189 + 195) / 9 = 165

Mean of the number of grunts:

(90 + 68 + 39 + 44 + 63 + 40 + 62 + 17 + 20) / 9 = 52

3. Subtract the mean of the age from each age value to get the deviation of each data point from the mean. Similarly, subtract the mean of the number of grunts from each number of grunts value.

Deviation of the age:

125 - 165 = -40

141 - 165 = -24

155 - 165 = -10

160 - 165 = -5

167 - 165 = 2

174 - 165 = 9

183 - 165 = 18

189 - 165 = 24

195 - 165 = 30

Deviation of the number of grunts:

90 - 52 = 38

68 - 52 = 16

39 - 52 = -13

44 - 52 = -8

63 - 52 = 11

40 - 52 = -12

62 - 52 = 10

17 - 52 = -35

20 - 52 = -32

4. Square each deviation value obtained in step 3.

Squared deviation of the age:

[tex](-40)^2 = 1600[/tex]

[tex](-24)^2 = 576[/tex]

[tex](-10)^2 = 100[/tex]

[tex](-5)^2 = 25[/tex]

[tex]2^2 = 4[/tex]

[tex]9^2 = 81[/tex]

[tex]18^2 = 324[/tex]

[tex]24^2 = 576[/tex]

[tex]30^2 = 900[/tex]

Squared deviation of the number of grunts:

[tex]38^2 = 1444[/tex]

[tex]16^2 = 256[/tex]

[tex](-13)^2 = 169[/tex]

[tex](-8)^2 = 64[/tex]

[tex]11^2 = 121[/tex]

[tex](-12)^2 = 144[/tex]

[tex]10^2 = 100[/tex]

[tex](-35)^2 = 1225[/tex]

[tex](-32)^2 = 1024[/tex]

5. Sum up all the squared deviation values obtained in step 4.

Sum of squared residuals:

1600 + 576 + 100 + 25 + 4 + 81 + 324 + 576 + 900 + 1444 + 256 + 169 + 64 + 121 + 144 + 100 + 1225 + 1024 = 8399

Therefore, the sum of the squared residuals of the least squared line for the given data is 8399.

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Answer:

a. 32,760 ways

b. 1365 ways

b. 1/1365

Step-by-step explanation:

Given

Number of qualified candidates= 15

Number of Vacancies = 4 ( President, CEO, COO and CFO)

a.

How many different ways can the officers be appointed?

This means that in how many ways can 4 candidates be arranged out of a total of 15.

They keyword here is arranged which means Permutation because the order of arrangement doesn't count.

So,

Number of Appointments = 15P4

Number of Appointment = 1365 ways

b. How many different ways can the committee be appointed?

Here, the order of appointments matters.

So, this means, in his many ways can a committee of 4 be selected out of a total of 15.

The keyword, selection means Combination.

So, number of Appointments = 15C4

Number of Appointments = 1365

c. What is the probability of randomly selecting the committee members and getting the five youngest of the qualified candidates?

There are 1365 ways of choosing the board members and the chance of selecting the 5 youngest members is 1, because there's only one way to select the 5 youngest.

So the probability = 1/1365

To appoint officers, there are 32,760 different ways. The committee can be appointed in 1,365 different ways. The probability of randomly selecting the four youngest candidates for the committee is approximately 0.073%.

To solve the problem of appointing corporate officers and a committee, we will go step by step. For part (a), there are 15 qualified candidates, and we need to appoint 4 different officers. Since one person cannot hold more than one officer position simultaneously, we will use permutations. The first officer can be chosen in 15 ways, the second in 14 ways, the third in 13 ways, and the fourth in 12 ways, making the total number of ways to appoint the officers:

15 × 14 × 13 × 12 = 32,760 different ways.

For part (b), after appointing the officers, 11 candidates remain (15 total candidates - 4 officers). However, since officers can also serve on the committee, we still have 15 candidates to choose from. We need to appoint 4 different committee members from these 15 candidates, which is a combination problem because the order in which they're chosen doesn't matter. The number of ways to appoint the committee is given by the combination formula C(n, k) = n! / (k!(n-k)!), where n is the total number of candidates, and k is the number of positions to fill:

C(15,4) = 15! / (4!(15-4)!) = 15! / (4! × 11!) = (15 × 14 × 13 × 12) / (4 × 3 × 2 × 1) = 1,365 different ways.

For part (c), to find the probability of randomly selecting the five youngest candidates of the 15 qualified candidates, we should note that there are 4 positions on the committee, not 5. Assuming 'five youngest' is a typo, the probability of getting the 4 youngest candidates on the committee is:

P(selecting 4 youngest) = 1 / C(15, 4) = 1 / 1,365 ≈ 0.00073, or 0.073%.

Answer:

The type 1 error here is a. Deciding that the absorption rates are different, when in fact they are not.

Step-by-step explanation:

A type I error is the rejection of a true null hypothesis (also known as a "false positive" finding or conclusion).

More generally, a Type I error occurs when a significance test results in the rejection of a true null hypothesis. By one common convention, if the probability value is below 0.05, then the null hypothesis is rejected.

In inferential statistics, the null hypothesis is a general statement or default position that there is nothing significantly different happening, like there is no association among groups or variables, or that there is no relationship between two measured phenomena.

Answer:

34.2022 < X < 39.7978

Step-by-step explanation:

For a 95% confidence interval, Z = 1.960

Sample size (n) = 33

Mean weight (X) = 37 ounces

Standard deviation (s) = 8.2 ounces

The relationship that describes a 95% confidence interval is:

[tex]X \pm 1.960*\frac{s}{\sqrt{n}}[/tex]

Applying the given data, the Lower (L) and Upper (U) limits are:

[tex]U=37 + 1.960*\frac{8.2}{\sqrt{33}} \\U=39.7978\\L=37 - 1.960*\frac{8.2}{\sqrt{33}} \\L=34.2022[/tex]

The 95% confidence interval is:

34.2022 < X < 39.7978

X is a discrete random variable.

A discrete random variable can only take on specific, distinct values with gaps in between.

In this case, X represents the number of imperfections in a computer chip, and it can only take on integer values: 0, 1, 2, 3, 4, or 5.

These values are countable and separate, indicating that X is a discrete random variable.

In contrast, a continuous random variable would have an infinite number of possible values within a range, and you would typically use intervals or real numbers to describe it.

For example, if we were measuring the weight of computer chips, it could be a continuous random variable because it could take on any value within a range, including fractions or decimals.

However, in this scenario, we are dealing with a countable and finite set of values for the number of imperfections, making X a discrete random variable.

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Answer: the store sold 31 1/3 pounds

Step-by-step explanation:

The store sold 15 2/3 pounds of carrots. Converting to improper fraction, it becomes 47/3 pounds.

The store sold 12 1/3 pounds of asparagus. Converting to improper fraction, it becomes 37/3 pounds.

The store sold 3 1/3 pounds of cabbage. Converting to improper fraction, it becomes 10/3 pounds.

Therefore, the total number of pounds that the store sold is

47/3 + 37/3 + 10/3 = 94/3 = 31 1/3 pounds

Answer:

[tex] E(Z) = E(X+Y) = E(X) +E(Y) = \mu_X +\mu_Y = 90+110=200[/tex]

[tex] Var(Z) = Var(X+Y) = Var(X) +Var(Y) +2 Cov(X,Y)[/tex]

Since X and Y are independent then [tex] Cov(X,Y) =0[/tex] and we have this:

[tex] Var(Z)= \sigma^2_X +\sigma^2_Y = 3^2 +4^2 = 9+16 =25[/tex]

And the deviation would be given by:

[tex] Sd(Z) = \sqrt{25}= 5[/tex]

And then the distribution for the total time would be given by:

[tex] Z= X+Y \sim N( \mu_Z= 200, \sigma_Z= 5)[/tex]

Step-by-step explanation:

For this case we can assume that X represent the flight time for the first filght and we know that:

[tex] X \sim N (\mu_X= 90. \sigma_x =3)[/tex]

And let Y the random variable that represent the time for the second filght and we know this:

[tex] Y \sim N(\mu_Y = 110, \sigma_Y =4)[/tex]

And we can define the random variable Z= X+Y as the total time for the two flights.

We can asume that X and Y are independent so then we have this:

[tex] E(Z) = E(X+Y) = E(X) +E(Y) = \mu_X +\mu_Y = 90+110=200[/tex]

[tex] Var(Z) = Var(X+Y) = Var(X) +Var(Y) +2 Cov(X,Y)[/tex]

Since X and Y are independent then [tex] Cov(X,Y) =0[/tex] and we have this:

[tex] Var(Z)= \sigma^2_X + \sigma^2_Y = 3^2 +4^2 = 9+16 =25[/tex]

And the deviation would be given by:

[tex] Sd(Z) = \sqrt{25}= 5[/tex]

And then the distribution for the total time would be given by:

[tex] Z= X+Y \sim N( \mu_Z= 200, \sigma_Z= 5)[/tex]

Answer:

0.267

Step-by-step explanation:

p = 0.3 q = 0.7

10C3 × p³ × q⁷

0.266827932

Answer:

26.68% probability that exactly three workers take public transportation daily

Step-by-step explanation:

For each worker, there are only two possible outcomes. Either they take public transportation daily, or they do not. The probability of a worker taking public transportation daily is independent from other workers. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

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

In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

And p is the probability of X happening.

30% of workers take public transportation daily.

This means that [tex]p = 0.3[/tex]

In a sample of 10 workers, what is the probability that exactly three workers take public transportation daily?

This is [tex]P(X = 3)[/tex] when [tex]n = 10[/tex]. So

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

[tex]P(X = 3) = C_{10,3}.(0.3)^{3}.(0.7)^{7} = 0.2668[/tex]

26.68% probability that exactly three workers take public transportation daily

Answer:

Step-by-step explanation:

given that after scoring a touchdown, a football team may elect to attempt a two-point conversion, by running or passing the ball into the end zone. If successful, the team scores two points.

X=1 if successful and

X=0 if not

pdf of X is

X      1       0

p    0.8   0.2

E(x) = [tex]1(0.8)+0(0.2)\\=0.8[/tex]

[tex]E(x^2) = 1^2(0.8) = 0.8[/tex]

Var(x) = 0.8-0.8*0.8

= 0.16

Now let us consider Y.

Y is the no of points scored.

Y      2       0

p     0.8    0.2

E(Y) = [tex]2(0.8)+0(0.2)\\=1,.6[/tex]

[tex]2^2(0.8)+0(0.2)\\= 3.2[/tex]

The mean and variance for X and Y are calculated using their definitions in probability theory. For X, the mean is 0.80 and variance is 0.16. For Y, the mean is 1.60 and variance is 0.64.

The random variable X is a Bernoulli random variable because it has only two outcomes: success (X=1) and failure (X=0). The mean and variance for a Bernoulli random variable can be found using the formulas: Mean (E[X]) = p and Variance (Var[X]) = p(1-p).

 For X:

The mean E[X] = p = 0.80. So, X = 1 with probability 0.80.

The variance Var[X] = p(1-p) = 0.80(1-0.80) = 0.16

The random variable Y represents the number of points scored which can either be 0 or 2. We let p be the probability of scoring 2 points i.e., p = 0.80. Hence, if the two-point conversion is successful, we will score 2 points, otherwise, we score 0.

For Y:

The mean E[Y] = 0*(1-p) + 2*p = 0 + 2*0.80 = 1.60

The variance Var[Y] = (0-E[Y])^2*(1-p) + (2-E[Y])^2*p = (0-1.60)^2*(1-0.80) + (2-1.60)^2*0.80 = 0.64

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Answer:

Step-by-step explanation:

Hello!

The researcher's objective is to compare the effectiveness of a water softener when used with a filtering process against its effectiveness when used without filtering.

To do so 90 locations were randomly divided into two equal groups.

Group A locations used the water softener with filtering.

Group B locations used the water softener without filtering.

At the end of three months, a water sample was taken of each location and its level of softness was registered (Scale 1 to 5, 5 represents the softest water)

X₁: Softness of water of a location from group A

n₁= 45 locations

X[bar]₁= 2.1

S₁= 0.7

X₂: Softness of water of a location from group B

n₂= 45 locations

X[bar]₂= 1.7

S₂= 0.4

To compare the effectiveness of the softener with and without a filtering process the parameter of interest is the difference between both population means:

Parameter: μ₁ - μ₂

The point estimation of the difference between the population means is the difference of the sample means: X[bar]₁ - X[bar]₂= 2.1-1.7= 0.4

Since the objective is to test if there is any difference with or without the filtering process, the hypothesis test to make is two-tailed:

H₀: μ₁ - μ₂ = 0

H₁: μ₁ - μ₂ ≠ 0

α: 0.05

Since there is no information about the distribution of both variables, you have to apply the central limit theorem and approximate the distribution of X[bar]₁ and X[bar]₂ to normal. Once both samples mean distribution is approximate to normal you can use the statistic:

[tex]Z= \frac{(X[bar]₁ - X[bar]₂)-(Mu_1-Mu_2)}{\sqrt{\frac{S_1^2}{n_1} +\frac{S_2^2}{n_2} } }[/tex]

[tex]Z_{H_0}= \frac{(2.1-1.7)-0}{\sqrt{\frac{0.49}{45} +\frac{0.16}{45} } } = 3.3282[/tex]

As said before, this test is two-tailed, so you will have two critical values:

Critical value 1: [tex]Z_{\alpha /2}= Z_{0.025}= -1.95[/tex]

Critical value 2: [tex]Z_{1-\alpha /2}= Z_{0.975}= 1.965[/tex]

The p-value of this test is also two tailed, you can calculate it as:

P(Z≥3.33) + P(Z≤3.33)= (1 - P(Z≤3.33))+P(Z≤-3.33)= (1-0.999566)+0.000434= 0.000868

p-value: 0.000868

This value means that 0.0868% of the sample size 45 taken from this population will provide natural evidence that there is no difference between the population means of the effectiveness of the water softener used with and without a filtering process.

A little reminder, the p-value is defined as the probability corresponding to the calculated statistic if possible under the null hypothesis (i.e. the probability of obtaining a value as extreme as the value of the statistic under the null hypothesis).

Both using the critical value method and the p-value method the decision is to reject the null hypothesis. This means that with a 5% level of significance there is a difference between the true average of the effectiveness of the water softener used with a filtering process and the true average effectiveness of the water softener used without a filtering process.

I hope it helps!

Answer:

Step-by-step explanation:

The original price of the wind chimes at the home improvement store is $w.

A customer signed up for a free membership card and received a 5% discount off the price. The value of the discount is

5/100 × w = 0.05w

The discounted price would be

w - 0.05w = 0.95w

Sales tax of 6% was applied after the discount. The amount of sales tax applied would be

6/100 × 0.95w = 0.057w

The algebraic expression to represent the final price of the wind chime is

0.95w + 0.057w

= 1.007w

Answer:

The probability of the stick's weight being 2.33 oz or greater is 0.41%.

Step-by-step explanation:

Test statistic (z) = (weight - mean)/sd

weight of stick = 2.33 oz

mean = 1.75 oz

sd = 0.22 oz

z = (2.33 - 1.75)/0.22 = 2.64

The cumulative area of the test statistic is the probability that the weight is 2.33 oz or less. The cumulative area is 0.9959.

The probability the weight is 2.33 or greater = 1 - 0.9959 = 0.0041 = 0.41%

Complete Question

The complete question is shown on the first uploaded image

Answer:

a

The Estimated  second point is (t,p) =(5,9320)

b

The rate is 180 thousand employer per year

c

The percentage rate of change of the function is 1.939%

Step-by-step explanation:

Looking at the graph

  First is to obtain the scale of the graph what i mean is what the distance between each line segment

    Considering the y-axis  each line segment is

                              [tex]\frac{9300-9200}{5} = \frac{100}{5} =20[/tex]

So this means that after 9300 the next line segment is 9320

Considering the x-axis each line segment is

                            [tex]\frac{4-2}{4} = \frac{2}{4} = 0.5[/tex]

What this means is that 2 line segment after 4 is 4 +2 ×(0.5) =5

So looking this two points (new_t,new_p) = (5 , 9320) = we see that they form a coordinate

       B) The labeled point that we are to consider are

   [tex](t_1,p_2) = (4.8, 9284) \ (t_2,p_2) = (5, 9320)[/tex]

The rate change

                 [tex]= \frac{p_2-p_1}{t_2-t_1}=\frac{9320-9284}{5-4.8} = \frac{36}{0.2} = 180[/tex]

So the rate is 180 thousand employer per year

C)

 So to obtain the percentage rate of change of the function

   Now

       [tex]f(4.8) = 9284 \ Thousand[/tex]

      [tex]f'(4.8) = 180 \ Thousand[/tex]    

Note: This is so because differentiation is the same as  slope of the graph

    Hence the percentage  rate of change

                          [tex]\frac{f'(4.8)}{f(4.8)} *\frac{100}{1} = \frac{180 \ 000}{9284 \ 000} * \frac{100}{1}[/tex]

                           = 1.939%

The question seems to be related to calculus, discussing concepts of tangents and rates of change for a function. To find another point on a tangent line or calculate the rate of change, we need the function and additional details. The respective formulas for these calculations are mentioned, but we can't provide a factual answer without more information.

Given the question, it seems this is related to the field of calculus, specifically addressing concepts of tangents and rates of change of a function. However, to find another point on the tangent line or calculate the rate of change at a specific point, we need more context or the actual function.

Usually, for the first question, you can use the formula slope = (y2-y1) / (x2-x1), assuming (t, p) is a point on the tangent and you know the slope of the tangent. For the second question, the rate of change at a point is the derivative of the function at that point. For the third question, the percentage rate of change at a point is the derivative at that point divided by the function value at that point, multiplied by 100 to get percentage.

Without the function and some additional information, we cannot arrive at a factual answer. You might want to check the question again and include the necessary details.

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Answer:

The probability is 0.4841.

Step-by-step explanation:

The provided table is:

From above table, it is known that

Number of subjects are 1008.

The probability that the person chosen is a woman given that the person is a light smoker can be calculated as:

[tex]P(Woman| Light smoker)=\frac{P(Woman and light smoker)}{P(Light smoker)} \\P(Woman| Light smoker)= \frac{\frac{76}{1008} }{\frac{157}{1008} } = 0.4841[/tex]

Thus, required probability is 0.4841.

The probability that a randomly selected person is a woman given that the person is a light smoker is [tex]\[ 0.400} \][/tex]

To find the probability that a randomly selected person is a woman given that the person is a light smoker, we need to use conditional probability. The formula for conditional probability [tex]\( P(A|B) \)[/tex] is:

[tex]\[P(A|B) = \frac{P(A \cap B)}{P(B)}\][/tex]

Where:

[tex]\( P(A|B) \)[/tex] is the probability of event [tex]\( A \)[/tex] occurring given that [tex]\( B \)[/tex] has occurred.

[tex]\( P(A \cap B) \)[/tex] is the probability of both events [tex]\( A \)[/tex] and [tex]\( B \)[/tex] occurring.

[tex]\( P(B) \)[/tex] is the probability of an event [tex]\( B \)[/tex] occurring.

Let's define the events:

[tex]\( A \)[/tex] : The person chosen is a woman.

[tex]\( B \)[/tex] : The person chosen is a light smoker.

We need the number of light smokers and the number of women who are light smokers. Suppose we have the following data:

Total number of subjects: 1008

Number of light smokers: [tex]\( n_{\text{light smokers}} \)[/tex]

A number of women who are light smokers: [tex]\( n_{\text{women and light smokers}} \)[/tex]

Given that the number of women who are light smokers is [tex]\( n_{\text{women and light smokers}} \)[/tex] and the total number of light smokers is [tex]\( n_{\text{light smokers}} \)[/tex], the probability can be calculated as follows:

[tex]\[P(\text{woman | light smoker}) = \frac{n_{\text{women and light smokers}}}{n_{\text{light smokers}}}\][/tex]

If we don't have the exact numbers, we'll need those to calculate the probability. However, let's assume the following values (hypothetically for the purpose of illustration):

Total number of light smokers: 150

Number of women who are light smokers: 60

The probability that a randomly selected person is a woman given that the person is a light smoker is:

[tex]\[P(\text{woman | light smoker}) = \frac{60}{150} = 0.4\][/tex]

Rounding to the nearest thousandth:

[tex]\[0.4 = 0.400\][/tex]