In analyzing hits by certain bombs in a​ war, an area was partitioned into 553 ​regions, each with an area of 0.95 km2. A total of 535 bombs hit the combined area of 553 regions. Assume that we want to find the probability that a randomly selected region had exactly two hits. In applying the Poisson probability distribution​ formula, ​P(x)equalsStartFraction mu Superscript x Baseline times e Superscript negative mu Over x exclamation mark EndFraction ​, identify the values of mu​, ​x, and e. ​Also, briefly describe what each of those symbols represents. Identify the values of mu​, ​x, and e.

Answer:

The total number of possible complete staffs you could hire is 350.

Step-by-step explanation:

The order that the software engineers are is not important. For example, hiring John and Laura as the software engineers is the same as hiring Laura and John. The same applies for the computer engineers. So we use the combinations formula to solve this question.

Combinations formula:

[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]

What is the total number of possible complete staffs you could hire?

2 software engineers from a set of 5

3 computer engineers from a set of 7

So

[tex]T = C_{5,2}*C_{7,3} = \frac{5!}{2!3!}*\frac{7!}{3!4!} = 10*35 = 350[/tex]

The total number of possible complete staffs you could hire is 350.

Answer:

The inspector's claim has strong statistical evidence.

Step-by-step explanation:

To answer this we have to perform a hypothesis test.

The inspector claimed that the actual proportion of code violations is greater than 0.07, so the null and alternative hypothesis are:

[tex]H_0: \pi\leq0.07\\\\H_a: \pi>0.07[/tex]

We assume a significance level of 0.05.

The sample size is 200 and the proportion of the sample is:

[tex]p=\frac{23}{200}= 0.115[/tex]

The standard deviation is

[tex]\sigma=\sqrt{\frac{\pi(1-\pi)}{N} }= \sqrt{\frac{0.07*0.93}{200}}=0.018[/tex]

The z-value can be calculated as

[tex]z=\frac{p-\pi-0.5/N}{\sigma} =\frac{0.115-0.07-0.5/200}{0.018} =\frac{0.0425}{0.018}=2.36[/tex]

The P-value for this z-value is P=0.00914.

This P-value is smaller than the significance level, so the effect is significant and the null hypothesis is rejected.

The inspector's claim has strong statistical evidence.

Answer:

Yes, because the p-value is 0.0062

Step-by-step explanation:

I looked it up on like 5 other websites and they all said this was the answer. I'm way too lazy to do it on my own.

A psychologist can use normative data, sequential design, or parent-report questionnaires to estimate the developmental score of a typical 4-year-old child without direct participation from children of that specific age.

To determine the developmental score of a typical 4-year-old child without having 4-year-old children participate in the study, a psychologist may use several statistical techniques grounded in developmental science. One method is to use normative data, which refers to average developmental milestones established through extensive research on children at various ages. By comparing existing data on children older and younger than 4 years, the psychologist can interpolate or approximate the developmental score for the typical 4-year-old based on these norms.

Another approach involves a sequential design, where overlapping cohorts of different age groups are studied over time. This method can help infer the developmental score of 4-year-olds by comparing changes in development across the different cohorts that are adjacent in age. Additionally, parent-report questionnaires can be used to gather data from parents or guardians about typical developmental milestones, which can then be statistically analyzed to estimate the developmental score of children in the age group of interest.

Answer:

Null Hypothesis: No first order autocorrelation

Alternative hypothesis: first order correlation exists

The assumptions to run this test are:

1) Errors are normally distributed with mean 0

2) Errors follows an stationary process

3) Independence condition between the erros

The statistic is defined as:

[tex] DW = \frac{\sum_{t=2}^ T (e_t -e_{t-1})^2}{\sum_{t=1}^T e^2_t}[/tex]

And if the value for the DW is near to 0 we can conclude that the assumption of Independence is not satisfied.

Step-by-step explanation:

The Durbin Watson test is a way to check autocorrelation in residuals for a time seeries or a regression.

We need to remember that the autocorrelation is the similarity of the time series in successive intervals. When we conduct this type of test we are checking if the time series can be modeled with and AR(1) process autoregressive.

The system of hypothesis on this case are:

Null Hypothesis: No first order autocorrelation

Alternative hypothesis: First order correlation exists

The assumptions to run this test are:

1) Errors are normally distributed with mean 0

2) Errors follows an stationary process

3) Independence condition between the errors

The statistic is defined as:

[tex] DW = \frac{\sum_{t=2}^ T (e_t -e_{t-1})^2}{\sum_{t=1}^T e^2_t}[/tex]

And if the value for the DW is near to 0 we can conclude that the assumption of Independence is not satisfied.

Answer:

a. P(tips from 60 parties > $600)=0.461

b. The total amount that she earns on the best 5% of such weekends is at least $19.43.

Step-by-step explanation:

a. To earn $600 in 60 parties means $10 per party in average.

If we assume a normal distribution of tips, we can calculate the z-value and its probability for this situation:

[tex]z=\frac{x-\mu}{\sigma}=\frac{10.00-9.40}{6.10}=\frac{0.60}{6.10}= 0.098\\\\P(x>10)=P(z>0.098)=0.461[/tex]

There is a probability of 46% that she earns at least $600 over a weekend of work.

b. The best 5% of the weekends corresponds to:

[tex]P(x>x_1)=0.05[/tex]

This probability (5%) corresponds to a z-value of z=1.6449.

In tips, this value represents:

[tex]x=\mu+z*\sigma=9.40+1.6449*6.10=9.40+10.03=19.43[/tex]

The total amount that she earns on the best 5% of such weekends is at least $19.43.

Final answer:

The probability of selecting the color of the first ball is 3/10. The probability of selecting the same color for the second ball depends on the remaining number of balls of that color divided by the total remaining number of balls.

Explanation:

The probability of selecting the color of the first ball is determined by the number of balls of that color divided by the total number of balls. In this case, there are 3 red balls out of 10, so the probability is 3/10.

After selecting the first ball without replacement, the probability of selecting the same color for the second ball depends on the remaining number of balls of that color divided by the total remaining number of balls. If the first ball was red, there are 2 red balls left out of 9, resulting in a probability of 2/9.

The probability of selecting a ball of the same color as the first ball drawn, without replacement, is as follows:

For red:[tex]\boxed{0.22}[/tex]

For white:[tex]\boxed{0.33}[/tex]

For blue:[tex]\boxed{0.22}[/tex]

Let's think step by step.To solve this problem, we need to calculate the probability of selecting a ball of the same color as the first ball drawn, without replacement. We will consider the two events separately: first, the probability of selecting a ball of the same color as the first ball when there are 10 balls in total, and second, the probability of selecting a ball of the same color as the first ball when there are only 9 balls left in the bucket (since the first ball is not replaced).

First Ball Selection:

When the first ball is picked, there are 10 balls in total, with 3 red, 4 white, and 3 blue balls. The probability of picking a ball of any specific color is the number of balls of that color divided by the total number of balls.

 For example, if the first ball drawn is red, the probability of drawing a red ball is:

[tex]\[ P(\text{first red}) = \frac{\text{Number of red balls}}{\text{Total number of balls}} = \frac{3}{10} \][/tex]

 Similarly, if the first ball is white, the probability is:

[tex]\[ P(\text{first white}) = \frac{\text{Number of white balls}}{\text{Total number of balls}} = \frac{4}{10} \][/tex]

 And if the first ball is blue, the probability is:

[tex]\[ P(\text{first blue}) = \frac{\text{Number of blue balls}}{\text{Total number of balls}} = \frac{3}{10} \][/tex]

 Second Ball Selection:

After the first ball is drawn and not replaced, there are 9 balls left in the bucket.

 If the first ball drawn was red, there are now 2 red balls left out of 9 total balls. The probability of drawing another red ball is:

[tex]\[ P(\text{second red} | \text{first red}) = \frac{\text{Number of red balls left}}{\text{Total number of balls left}} = \frac{2}{9} \][/tex]

If the first ball was white, there are now 3 white balls left out of 9 total balls. The probability of drawing another white ball is:

[tex]\[ P(\text{second white} | \text{first white}) = \frac{\text{Number of white balls left}}{\text{Total number of balls left}} = \frac{3}{9} = \frac{1}{3} \][/tex]

If the first ball was blue, there are now 2 blue balls left out of 9 total balls. The probability of drawing another blue ball is:

[tex]\[ P(\text{second blue} | \text{first blue}) = \frac{\text{Number of blue balls left}}{\text{Total number of balls left}} = \frac{2}{9} \][/tex]

Now, let's calculate the probabilities for each color and round them to two decimal places:

 For red:

[tex]\[ P(\text{second red} | \text{first red}) = \frac{2}{9} \approx 0.22 \][/tex]

 For white:

[tex]\[ P(\text{second white} | \text{first white}) = \frac{1}{3} \approx 0.33 \][/tex]

 For blue:

[tex]\[ P(\text{second blue} | \text{first blue}) = \frac{2}{9} \approx 0.22 \][/tex]

Final

The probability of selecting a ball of the same color as the first ball drawn, without replacement, is as follows:

For red:[tex]\boxed{0.22}[/tex]

For white:[tex]\boxed{0.33}[/tex]

For blue:[tex]\boxed{0.22}[/tex]

 These probabilities are rounded to two decimal places.

 The answer is: 0.22.

÷Answer:

Standard Deviation  =  176.5

Step-by-step explanation:

To calculate the standard deviation, calculate the mean score for the 4 standard deviation scores:

mean, m = Σx ÷ n

where Σx represents summation of each value = 162 + 153 + 317 + 413

= 1045

n = number of samples to be considered = 4

mean, m = 1045 ÷4

= 261.25

To calculate the standard deviation, use the formula below

SD    =   [tex]\sqrt{\frac{Σ(x-m)}{n} ^{2} }[/tex]

where x =  each value from the week lead time

m = mean = 261.25

n = the size = 4

The Standard deviation formula can be simplified further

when x = 162

[tex]\sqrt{\frac{(x1-m)}{n} ^{2} }[/tex] = 49.625

when x = 153

[tex]\sqrt{\frac{(x2-m)}{n} ^{2} }[/tex] = 23.125

when x = 317

[tex]\sqrt{\frac{(x3-m)}{n} ^{2} }[/tex]=  27.875

when x = 413

[tex]\sqrt{\frac{(x4-m)}{n} ^{2} }[/tex]= 75.875

Note that the above 4 equations can be lumped up into one giant equation by applying a big square root function instead of breaking it down

SD = 49.625 + 23.125 + 27.875 + 75.875

SD = 176.5

QUESTION IS INCOMPLETE.

Nevertheless, I will explain how to find, without solving, the longest interval in which an initial value problem is certain to have a unique twice differentiable solution.

Step-by-step explanation:

Consider the Existence and uniqueness theorem:

Let p(t) , q(t) and r(t) be continuous on an interval a ≤ t ≤ b, then the differential equation given by:

y''+ p(t) y' +q(t) y = r(t) ;

y(t_0) = y_0, y'(t_0) = y'_0

has a unique solution defined for all t in the stated interval.

Example:

Consider the differential equation

ty'' + 9y = t

y(1) = y_0,

y'(1) = v_0

ty'' + 9y = t .................................(1)

First, write the differential equation (1) in the form:

y'' + p(t)y' + q(t)y = r(t) ..................(2)

by dividing (1) by t

So

y''+ (9/t)y = 1 ....................................(3)

Comparing (3) with (2)

p(t) = 0

q(t) = 9/t

r(t) = 1

For t = 0, p(t) and r(t) are continuous, but q(t) is undefined.

q(t) is continuous everywhere apart from the point t = 0.

We say (-∞, 0) and (0,∞) are the points where p(t), q(t) and r(t) are continuous.

But t = 1, which is contained in the initial conditions y(1) = y_0 and y'(1) = v_0 is found in (0,∞).

So, we conclude that this interval is the longest interval in which the initial value problem has a unique twice differentiable solution.

Answer:

[tex]0.322 - 1.88\sqrt{\frac{0.322(1-0.322)}{500}}=0.283[/tex]

[tex]0.322 + 1.88\sqrt{\frac{0.322(1-0.322)}{500}}=0.361[/tex]

The 94% confidence interval would be given by (0.283;0.361)

Step-by-step explanation:

Notation and definitions

[tex]n=500[/tex] random sample taken

[tex]\hat p=0.322[/tex] estimated proportion of people between 18 to 34 who live with their parents

[tex]p[/tex] true population proportion of people between 18 to 34 who live with their parents  

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".  

The population proportion have the following distribution

[tex]p \sim N(p,\sqrt{\frac{p(1-p)}{n}})[/tex]

Solution to the problem

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 94% of confidence, our significance level would be given by [tex]\alpha=1-0.94=0.06[/tex] and [tex]\alpha/2 =0.03[/tex]. And the critical value would be given by:

[tex]z_{\alpha/2}=-1.88, z_{1-\alpha/2}=1.88[/tex]

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

[tex]\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex]

If we replace the values obtained we got:

[tex]0.322 - 1.88\sqrt{\frac{0.322(1-0.322)}{500}}=0.283[/tex]

[tex]0.322 + 1.88\sqrt{\frac{0.322(1-0.322)}{500}}=0.361[/tex]

The 94% confidence interval would be given by (0.283;0.361)

Answer:

Research hypothesis

Step-by-step explanation:

A specific, clear, and testable proposition or predictive statement about the possible outcome of a scientific research study based on a particular property of a population, such as presumed differences between groups on a particular variable or relationships between variables is known as a research hypothesis  .

One of the most important steps in planning a scientific quantitative research study is specifying the research hypotheses. A prior expectation about the results of the study in one or more research hypotheses  is usually being stated by a  quantitative researcher before conducting the study, because the design of the research study and the planned research design is often determined by the stated hypotheses.

In the question, it is clearly seen that a prior expectation about the results of the study on the relationship between income and self-esteem was already suggested, before been tested. Thus, it is a research hypothesis

Final answer:

To find the probability that the size of the family is between 2 and 5 inclusive, calculate the proportion of the given sample that falls within that range. In this case, the probability is approximately 0.792.

Explanation:

To find the probability that the size of the family is between 2 and 5 inclusive, we need to calculate the proportion of the given sample that falls within that range. In the provided sample of college math class, the family sizes are:

We can see that there are 19 family sizes falling between 2 and 5, inclusive. Therefore, the probability is:

Probability = Number of favorable outcomes / Total number of outcomes

Probability = 19 / 24 = 0.792

So, the probability that the size of the family is between 2 and 5 inclusive is approximately 0.792.

Answer:

a) The 90% confidence interval is:

[tex]0.0003237 \leq \pi \leq 0.0003239[/tex]

b) Yes. Because the proportion of cancer rate for the cell phone users population is bigger than 0.00324%, which is over the rate of non-cell phone users (0.0251%).

Step-by-step explanation:

a) We will construct the 90% confidence interval based on the information given by the sample taken in this study.

The sample proportion is:

[tex]p=\frac{X}{n}=\frac{136}{420,026} =0.000324[/tex]

The standard deviation is estimated as:

[tex]\sigma=\sqrt{\frac{p(1-p)}{n}}= \sqrt{\frac{0.000324*0.999676}{420,026}}=\sqrt{7.7\cdot 10^{-10}} = 0.000028[/tex]

As the sample size is big enough, we use the z-score. For a 90% CI, the value of z is z=1.645.

The margin of error of the CI is:

[tex]E=z\sigma/\sqrt{n}=1.645* 0.000028 /\sqrt{420,026}\\\\E= 0.000046 /648= 0.00000007[/tex]

The 90% CI is:

[tex]p-z\sigma/\sqrt{n}\leq \pi\leq p+z\sigma/\sqrt{n}\\\\0.000324- 0.00000007 \leq\pi\leq 0.000324+ 0.00000007\\\\ 0.0003237 \leq \pi \leq 0.0003239[/tex]

Final Answer:

a. The 90% confidence interval estimate of the percentage of cell phone users who develop cancer of the brain or nervous system, rounded to three decimal places, is approximately: [0.028%, 0.037%]
b. The known rate for those not using cell phones is 0.0251%.

Explanation:

a. To construct a 90% confidence interval estimate of the percentage of cell phone users who develop cancer of the brain or nervous system, we need to know the number of successes (in this case, the number of people who developed cancer), the total number of trials (the total number of cell phone users), and use the z-score that corresponds to the desired confidence level (90%).
Given:
Number of cell phone users (n) = 420,026
Number of cell phone users who developed cancer (x) = 136
Confidence level (CL) = 90%

First, we calculate the sample proportion (p):
p = x / n = 136 / 420,026 ≈ 0.000324

For a 90% confidence interval, the z-score associated with the two-tailed confidence level is approximately 1.645 (you would find this value in a z-score table or by using statistical software).

Next, we calculate the standard error (SE) for the proportion:
SE = sqrt((p(1 - p)) / n)
SE = sqrt((0.000324(1 - 0.000324)) / 420,026)
SE ≈ sqrt((0.000324 * 0.999676) / 420,026)
SE ≈ sqrt(0.00000032443 / 420,026)
SE ≈ sqrt(7.72476e-10)
SE ≈ 2.779e-5

Now, we calculate the margin of error (ME):
ME = z * SE
ME = 1.645 * 2.779e-5
ME ≈ 4.570635e-5

Finally, we construct the confidence interval:
Lower bound (LB) = p - ME
LB ≈ 0.000324 - 4.570635e-5
LB ≈ 0.000278

Upper bound (UB) = p + ME
UB ≈ 0.000324 + 4.570635e-5
UB ≈ 0.000370

So the 90% confidence interval estimate of the percentage of cell phone users who develop cancer of the brain or nervous system, rounded to three decimal places, is approximately:
[0.028%, 0.037%]

b. To determine if cell phone users appear to have a rate of cancer that is different from the rate among those not using cell phones, we compare the confidence interval we just calculated with the known rate for non-cell phone users.
The known rate for those not using cell phones is 0.0251%.

Since the entire 90% confidence interval of [0.028%, 0.037%] for cell phone users is above 0.0251%, this suggests that the rate of cancer among cell phone users might be higher than the rate among non-users. However, this does not provide enough evidence to definitively conclude that cell phone use causes a higher rate of cancer since other factors might be at play and causation cannot be determined from this data alone. It is also important to note that the confidence interval is statistically very close to the known rate for non-users, so the difference might not be practically significant. Additionally, further research with more rigorous control conditions would be necessary to establish a cause-and-effect relationship.

Answer:

a) Probability of no broken bats in a game P(X=0) = 0.3678

b) Probability of at-least broken bats in a game P(X≥2) = 0.2644

Step-by-step explanation:

we will use Poisson distribution

P(X=x) = e^(-λ) λ^ r/r!

a) Probability of no broken bats in a game P(X=0) = e^-1

                                                                                   = 0.3678

b) Probability of at-least two broken bats in a game

                                                                     P(X≥2) = 1-([p(x=0)+P(x=1)]

                                                                               = 1- (e^(-1)+e^(-1))

                                                                                  = 1-(0.3678+0.3678)

                                                                   P(X≥2)= 1- 0.7356

                                                                  P(X≥2)= 0.2644

Answer:

The standard deviation is 5.83 kg

Step-by-step explanation:

Variance and Standard Deviation

Given a data set of random values, the variance is defined as the average of the squared differences from the mean. We use this formula to calculate the variance:

[tex]\displaystyle \sigma^2=\frac{\sum(x_i-\mu)^2}{n}[/tex]

Where [tex]\mu[/tex] is the mean of the values xi (i=1 to n), n the total number of values.

The standard deviation is known by the symbol [tex]\sigma[/tex] and is the square root of the variance.

We are given the value of the variance:

[tex]\sigma^2=34\ kg^2[/tex]

We now compute the standard deviation

[tex]\sigma=\sqrt{34\ kg^2}=5.83\ kg[/tex]

The standard deviation is 5.83 kg

Answer:

a.

[tex]\bar p_1=0.05\\\bar p_2=0.067[/tex]

b-Check illustration  below

c.(-0.0517,0.0177

Step-by-step explanation:

a.let [tex]p_1 \& p_2[/tex] denote processes 1 & 2.

For [tex]p_1[/tex]: T1=10,n1=200

For [tex]p_2[/tex]:T2=20,n2=300

Therefore

[tex]\bar p_1=\frac{t_1}{N_1}=\frac{10}{200}=0.05\\\bar p_2=\frac{t_2}{N_2}=\frac{20}{300}=0.067[/tex]

b. To test for hypothesis:-

i.

[tex]H_0:p_1=p_2\\H_A=p_1\neq p_2\\\alpha=0.05[/tex]

ii.For a two sample Proportion test

[tex]Z=\frac{\bar p_1-\bar p_2}{\sqrt(\bar p(1-\bar p)(\frac{1}{n_1}+\frac{1}{n_2})}\\[/tex]

iii. for [tex]\frac{\alpha}{2}=(-1.96,+1.96)[/tex] (0.5 alpha IS 0.025),

reject [tex]H_o[/tex] if[tex]|Z|>1.96[/tex]

iv. Do not reject [tex]H_o[/tex]. The noncomforting proportions are not significantly different as calculated below:

[tex]z=\frac{0.050-0.067}{\sqrt {(0.06\times0.94)\times \frac{1}{500}}}[/tex]

z=-0.78

c.[tex](1-\alpha).100\%[/tex] for the p1-p2 is given as:

[tex](\bar p_1-\bar p_2)\pm Z_0_._5_\alpha \times \sqrt \frac{ \bar p_1(1-\bar p_1)}{n_1}+\frac{\bar p_2(1-\bar p_2)}{n_2}\\\\=(0.05-0.067)\pm 1.645 \times \sqrt \ \frac{0.05+0.95}{200}+\frac{0.067+0.933}{300}\\[/tex]

=(-0.0517,+0.0177)

*CI contains o, which implies that proportions are NOT significantly different.

Answer and Step-by-step explanation:

Obviously addition is closed for R including both infinities

As t+infty = infty +t = infty and t+(-infty) =(-infty)+t =(-infty)

and inverse are -infty for infty and infty for -infty

Hence a group under addition

------------------------------------------------------------

But regarding multiplication

we can say 1/infty = 2/infty =0

Hence infinity*0 is not unique.

Also infinity and -infty do not have multiplicative inverse as there is no t such that t*infty = 1

Hence cannot be a vector space.

Answer:

(a) Customer will not purchase the light bulbs at significance level of 0.05

(b) Customer will purchase the light bulbs at significance level of 0.01 .

Step-by-step explanation:

We are given that Light bulbs of a certain type are advertised as having an average lifetime of 750 hours. A random sample of 50 bulbs was selected,  and the following information obtained:

Average lifetime = 738.44 hours and a standard deviation of lifetimes = 38.2 hours.

Let Null hypothesis, [tex]H_0[/tex] : [tex]\mu[/tex] = 750 {means that the true average lifetime is same as what is advertised}

Alternate Hypothesis, [tex]H_1[/tex] : [tex]\mu[/tex] < 750 {means that the true average lifetime is smaller than what is advertised}

Now, the test statistics is given by;

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

where, X bar = sample mean = 738.44 hours

               s  = sample standard deviation = 38.2 hours

               n = sample size = 50

So, test statistics = [tex]\frac{738.44-750}{\frac{38.2}{\sqrt{50} } }[/tex] ~ [tex]t_4_9[/tex]

                            = -2.14

(a) Now, at 5% significance level, t table gives critical value of -1.6768 at 49 degree of freedom. Since our test statistics is less than the critical value of t, so which means our test statistics will lie in the rejection region and we have sufficient evidence to reject null hypothesis.

Therefore, we conclude that the true average lifetime is smaller than what is advertised and so consumer will not purchase the light bulbs.

(b) Now, at 1% significance level, t table gives critical value of -2.405 at 49 degree of freedom. Since our test statistics is higher than the critical value of t, so which means our test statistics will not lie in the rejection region and we have insufficient evidence to reject null hypothesis.

Therefore, we conclude that the true average lifetime is same as what it has been advertised and so consumer will purchase the light bulbs.

Answer:

6 Full Time Employees

Step-by-step explanation:

It takes 20 minutes to code each record.

Therefore, in 1 hour, number of records per coder = 60/20 = 3 records

Since each coding professional works 7.5 productive hours per day.

Number of Records Per Day for each coder

=(3X7.5) = 22.5 records

Assume a 5-Week Day

Number of Records that must be treated = 600/5 = 120 records a day

Number of FTE Employees needed

= 120 / 22.5 = 5.33

The employer will require at least 6 FTE since Number of Employees is a discrete data.

Final answer:

The coding supervisor will need approximately 5.33 FTEs to code 600 discharges per week, considering the given average time to code each record and daily productive hours. Since a fraction of an FTE isn't practical, the supervisor will need to employ 6 full-time coding professionals.

Explanation:

To determine the number of Full-Time Equivalents (FTEs) needed to code 600 discharges per week, we first need to calculate the total time required to code all records and then divide that time by the productive hours each coding professional works per day.

First, calculate the total weekly coding time: 600 discharges × 20 minutes per discharge = 12,000 minutes per week.

Then convert the total weekly coding time to hours: 12,000 minutes / 60 = 200 hours per week needed for coding.

Considering each coding professional works 7.5 productive hours per day, let's find out the daily capacity for one FTE: 7.5 hours per day × 5 days per week = 37.5 hours per week.

Now, divide the total weekly hours needed by the weekly capacity of one FTE: 200 hours per week / 37.5 hours per FTE per week = approximately 5.33 FTEs.

Therefore, the coding supervisor will need to employ roughly 5.33 FTEs to code 600 discharges per week, given the average coding time and productive hours per day. Since you can't have a fraction of an FTE in reality, the supervisor would need to round up to employ 6 full-time coding professionals.

Answer:

-8 - 3i

Step-by-step explanation:

To find the Complex Conjugate , Negate the term with i

Hopefully this helps.

Answer:

a.

[tex]P(X=3)=0.2903\\\\P(X \geq 3)=0.5801\\\\P(X\leq 3)0.7102[/tex]

b.

[tex]P(2\leq x\leq 4 )=0.7451[/tex]

c. mean=2.8

d . standard deviation=1.2961

Step-by-step explanation:

We determine that the accident rates follow a binomial distribution. The rate of success p=0.4 and sample n=7:

[tex]P(x)={n\choose x}p^x(1-p)^{n-x}[/tex]

#the probability of exactly​ three;

[tex]P(x)={n\choose x}p^x(1-p)^{n-x}\\\\P(X=3)={7\choose 3}0.4^3(0.6)^{4}\\\\=0.2903[/tex]

#At least(more than 2)

[tex]P(x)={n\choose x}p^x(1-p)^{n-x}\\\\P(X\geq 3)=1-P(X\leq 2)\\\\=1-{7\choose 0}0.4^0(0.6)^{7}-{7\choose 1}0.4^1(0.6)^{6}-{7\choose 2}0.4^2(0.6)^{5}\\\\=1-0.0280-0.1306-0.2613\\\\=0.5801[/tex]

#At most 3;

[tex]P(x)={n\choose x}p^x(1-p)^{n-x}\\\\P(X \leq 3)={7\choose 0}0.4^0(0.6)^{7}+{7\choose 1}0.4^1(0.6)^{6}+{7\choose 2}0.4^2(0.6)^{5}+{7\choose 3}0.4^3(0.6)^{4}\\\\\\\=0.0280+0.1306+0.2613+0.2903\\\\=0.7102[/tex]

b. Between 2 and 4:

Using the binomial expression, this probability is calculated as:

[tex]P(x)={n\choose x}p^x(1-p)^{n-x}\\\\P(2\leq x\leq 4 )={7\choose 2}0.4^2(0.6)^{5}+{7\choose 3}0.4^3(0.6)^{4}+{7\choose 4}0.4^4(0.6)^{3}\\\\\\\\\=0.2613+0.2903+0.1935\\\\=0.7451[/tex]

Hence,the probability of between 2 and four is 0.7451

c. From a above, we have the values of n=7 and p=0.4.

-We substitute this values in the formula below to calculate the mean:

-The mean of a binomial distribution is calculated as the product of the probability of success by the sample size, mean=np:

[tex]\mu=np, n=7, p=0.4\\\\\mu=7\times 0.4\\\\=2.8[/tex]

Hence, the standard deviation of the sample is 2.8

d. From a above, we have the values of n=7 and p=0.4

--We substitute this values in the formula below to calculate the standard deviation

-The standard deviation a binomial distribution is given as:

[tex]\sigma={\sqrt {np(1-p)}\\\\=\sqrt{7\times 0.4\times 0.6}\\\\=1.2961[/tex]

Hence, the standard deviation of the sample is 1.2961

We can calculate the probabilities of having exactly 3 fatalities, at least 3 fatalities, and at most 3 fatalities using the formula. The mean and standard deviation of the random variable Y can be calculated using specific formulas for a binomial distribution.

To solve this problem, we will use the binomial probability formula. The probability of exactly 3 traffic fatalities involving an intoxicated or alcohol-impaired driver or nonoccupant can be calculated by plugging in the values of n (total number of trials) and p (probability of success in a single trial) into the formula. The probability of at least 3, or more than 3 traffic fatalities involving an intoxicated or alcohol-impaired driver or nonoccupant can be calculated by summing the probabilities of 3, 4, 5, 6, and 7 fatalities. The probability of at most 3 traffic fatalities involving an intoxicated or alcohol-impaired driver or nonoccupant can be calculated by summing the probabilities of 0, 1, 2, and 3 fatalities. The mean of the random variable Y can be calculated by multiplying the number of trials (7) by the probability of success (0.4). The standard deviation of Y can be calculated using the formula for the standard deviation of a binomial distribution.

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Answer:

1.83% probability there are no car accidents on that stretch on Monday

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]

In which

x is the number of sucesses

e = 2.71828 is the Euler number

[tex]\mu[/tex] is the mean in the given time interval.

The number of accidents on a certain section of I-40 averages 4 accidents per weekday independent across weekdays.

This means that [tex]\mu = 4[/tex]

What is the probability there are no car accidents on that stretch on Monday?

This is P(X = 0).

[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]

[tex]P(X = 0) = \frac{e^{-4}*(4)^{0}}{(0)!} = 0.0183[/tex]

1.83% probability there are no car accidents on that stretch on Monday

The probability of there being no car accidents on Monday on a certain section of I-40 can be calculated using the Poisson distribution. In this case, with an average of 4 accidents per weekday, the probability is approximately 1.83%.

To calculate the probability of there being no car accidents on Monday, we can use the Poisson distribution. In this case, the average number of accidents per weekday is given as 4. The Poisson distribution can be used to calculate the probability of a specific number of events occurring in a given time period.

The formula for calculating the probability of x events occurring in a Poisson distribution is:

P(x) = (e^-λ * λ^x) / x!

Where λ is the average number of events, and x is the number of events we want to calculate the probability for.

In this case, the average number of accidents per weekday is 4, so λ = 4. And we want to calculate the probability of there being no accidents, so x = 0.

Using the formula, we can calculate:

Therefore, the probability of there being no car accidents on Monday is approximately 0.0183, or 1.83%.

Answer:

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

[tex]X \sim N(\mu,\sigma)[/tex]  

Where [tex]\mu[/tex] the mean and [tex]\sigma[/tex]  the deviation

We know that the z score is given by:

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

And by properties the value that separate the half  area on one side and half is on the other is z=0, since we have this:

[tex] P(Z<0) =0.5[/tex]

[tex] P(Z>0)=0.5[/tex]

So then the correct answer for this case would be z =0.00

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

[tex]X \sim N(\mu,\sigma)[/tex]  

Where [tex]\mu[/tex] the mean and [tex]\sigma[/tex]  the deviation

We know that the z score is given by:

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

And by properties the value that separate the half  area on one side and half is on the other is z=0, since we have this:

[tex] P(Z<0) =0.5[/tex]

[tex] P(Z>0)=0.5[/tex]

So then the correct answer for this case would be z =0.00

The z-score that divides the standard normal distribution into two equal halves is 0.00 as the mean of this distribution is also 0. Z-scores are standardized values expressing how many standard deviations a value is above or below the mean and can be determined using a Z-Table of Standard Normal Distribution.

The value of z that divides the standard normal distribution so that half the area is on one side and half is on the other is 0.00. The standard normal distribution is a symmetrical distribution where half of the values are less than the mean and half of the values are greater than the mean. In a standard normal distribution, the mean is 0. Hence, the z-score that will divide the distribution into half is also 0.

A z-score is a standardized value, expressing how many standard deviations a value is above or below the mean. These scores are particularly useful when comparing values from different data sets that have different means and standard deviations, such as exam scores from different classes or schools. They help us understand whether a particular score is common, or exceptionally high or low.

To find this value, you can use a Z-Table of Standard Normal Distribution, which shows the cumulative probability of a random variable from a standard normal distribution (with mean 0 and standard deviation 1), as being less than a certain value.

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Answer:

(a) Null hypothesis: The new drug reduces the rate of first heart attacks by a percentage same as aspirin.

(b) Alternate hypothesis: The new drug reduces the rate of first heart attacks by a percentage greater than aspirin.

Step-by-step explanation:

(a) A null hypothesis is a statement from the population parameter which is either rejected or accepted (fail to reject) upon testing. It is expressed using the equality sign.

(b) An alternate hypothesis is also a statement from the population parameter which negates the null hypothesis and is accepted if the null hypothesis is rejected. It is expressed using any of the inequality signs.

Answer:

[tex](a) = \frac{144}{133225} \\\\(b) = \frac{1}{365}[/tex]

Step-by-step explanation:

Part (a) the probability that two people have a birthday on the 9th of any month.

Neglecting leap year, there are 365 days in a year.

There are 12 possible 9th in months that make a year calendar.

If two people have birthday on 9th; P(1st person) and P(2nd person).

[tex]=\frac{12}{365} X\frac{12}{365} = \frac{144}{133225}[/tex]

Part (b) the probability that two people have a birthday on the same day of the same month

P(2 people selected have birthday on the same day of same month) + P(2 people selected not having birthday on  same day of same month) = 1

P(2 people selected not having birthday on  same day of same month):

[tex]= \frac{365}{365} X \frac{364}{365} =\frac{364}{365}[/tex]

P(2 people selected have birthday on the same day of same month) [tex]= 1-\frac{364}{365} \\\\= \frac{1}{365}[/tex]

Final answer:

The probability that two people have a birthday on the 9th of any month is 1/133,225. The probability that two people have a birthday on the same day of the same month is also 1/133,225.

Explanation:

To calculate the probability that two people have a birthday on the 9th of any month, we need to consider the number of possible outcomes and the number of favorable outcomes. There are 12 months in a year, so the number of possible outcomes is 12. The probability of each person having a birthday on the 9th is 1/365. Therefore, the probability that two people have a birthday on the 9th of any month is (1/365) x (1/365) = 1/133,225.

To calculate the probability that two people have a birthday on the same day of the same month, we need to consider the number of possible outcomes and the number of favorable outcomes. There are 12 months in a year and each month has 30 or 31 days. So the number of possible outcomes is 12 x 31 = 372. The probability of each person having a birthday on a specific day is 1/365. Therefore, the probability that two people have a birthday on the same day of the same month is (1/365) x (1/365) = 1/133,225.

Answer:

In a sample of 1000 test the expected number of false negatives is 8.6.

Step-by-step explanation:

The table provided is:

                             Positive      Negative    TOTAL

Hepatitis C             335                  10            345

No Hepatitis C           2               1153           1155

TOTAL                     337               1163          1500

A false negative test result implies that, the person's test result for Hepatitis C was negative but actually he/she had Hepatitis C.

Compute the probability of false negative as follows:

[tex]P(Negative|No\ Hepapatits\ C)=\frac{n(Negative\cap No\ Hepapatits\ C)}{n(No\ Hepapatits\ C)} \\=\frac{10}{1163}\\ =0.0086[/tex]

Compute the expected number of false negatives in a sample is n = 1000 tests as follows:

E (False negative) = n × P (Negative|No Hepatitis C)

                              [tex]=1000\times0.0086\\=8.6[/tex]

Thus, in a sample of 1000 test the expected number of false negatives is 8.6.

Answer:

90 percent confidence interval = [72.674 ,77.326]

Step-by-step explanation:

We are given that weight of a product is measured in pounds.

A random sample of 50 units is taken from a recent production. The sample yielded y bar = 75 lb, and we know that [tex]\sigma^{2}[/tex] = 100 lb.

The Pivotal quantity for 9% confidence interval is given by;

              [tex]\frac{Ybar - \mu}{\frac{\sigma}{\sqrt{n} } }[/tex] ~ N(0,1)

where, Y bar = sample mean = 75

                [tex]\sigma[/tex]  = population standard deviation = 10

                 n = sample size = 50

So, 90% confidence interval for population mean,  is given by;

P(-1.6449 < N(0,1) < 1.6449) = 0.90

P(-1.6449 < [tex]\frac{Ybar - \mu}{\frac{\sigma}{\sqrt{n} } }[/tex] < 1.6449) = 0.90

P(-1.6449 * [tex]{\frac{\sigma}{\sqrt{n} }[/tex] < [tex]{Ybar - \mu}[/tex] < 1.6449 * [tex]{\frac{\sigma}{\sqrt{n} }[/tex] ) = 0.90

P(Y bar - 1.6449 * [tex]{\frac{\sigma}{\sqrt{n} }[/tex] < [tex]\mu[/tex] < Y bar + 1.6449 * [tex]{\frac{\sigma}{\sqrt{n} } }[/tex] ) = 0.90

90% confidence interval for [tex]\mu[/tex] = [ Y bar - 1.6449 * [tex]{\frac{\sigma}{\sqrt{n} }[/tex] , Y bar + 1.6449 * [tex]{\frac{\sigma}{\sqrt{n} }[/tex] ]

                                              = [ 75 - 1.6449 * [tex]{\frac{10}{\sqrt{50} }[/tex] , 75 + 1.6449 * [tex]{\frac{10}{\sqrt{50} }[/tex] ]

                                              = [ 72.674 , 77.326 ]

Therefore, 90% confidence interval for population mean is [72.674 ,77.326] .

Final answer:

To calculate how long it will take to double an investment with a 5.5% annual interest rate, you can use the Rule of 72, which gives you approximately 13.09 years for doubling the investment.

Explanation:

To determine how long it will take to double your money with an annual interest rate of 5.5%, you can use the Rule of 72. The Rule of 72 is a simple formula to estimate the number of years required to double the invested money at a given annual fixed interest rate. You divide 72 by the annual interest rate.

In this situation, divide 72 by 5.5 to find out how many years it will take to double the investment:

72 / 5.5 = 13.09 years

Thus, it will take approximately 13.09 years to double your money at an annual interest rate of 5.5%, when the interest is compounded annually. This is a rough approximation and the actual number might vary slightly depending on the compounding method used by the account.

Answer:

26% probability that they win both contracts.

Step-by-step explanation:

These following probabilities are important to solve this question:

0.4 = 40% probability of winning the first contract.

0.65 = 65% probability of winning the second contract if the first contract is won.

What is the probability that they win both contracts?

[tex]P = 0.4*0.65 = 0.26[/tex]

26% probability that they win both contracts.

Answer:

[tex]P(5<X<7)=P(\frac{5-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{7-\mu}{\sigma})=P(\frac{5-6}{0.5}<Z<\frac{7-6}{0.5})=P(-2<z<2)[/tex]

And we can find this probability with thie difference:

[tex]P(-2<z<2)=P(z<2)-P(z<-2)[/tex]

And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.  

[tex]P(-2<z<2)=P(z<2)-P(z<-2)=0.97725-0.02275=0.9545[/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 heights of a population, and for this case we know the distribution for X is given by:

[tex]X \sim N(6,0.5)[/tex]  

Where [tex]\mu=6[/tex] and [tex]\sigma=0.5[/tex]

We are interested on this probability

[tex]P(5<X<7)[/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]

If we apply this formula to our probability we got this:

[tex]P(5<X<7)=P(\frac{5-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{7-\mu}{\sigma})=P(\frac{5-6}{0.5}<Z<\frac{7-6}{0.5})=P(-2<z<2)[/tex]

And we can find this probability with thie difference:

[tex]P(-2<z<2)=P(z<2)-P(z<-2)[/tex]

And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.  

[tex]P(-2<z<2)=P(z<2)-P(z<-2)=0.97725-0.02275=0.9545[/tex]

Final answer:

The probability that a 44 inch television screen will have a service life between 5 and 7 years, given a normal distribution with a mean of 6 years and a standard deviation of 0.5 years, is approximately 95.45%.

Explanation:

To calculate the probability that the service life of a 44 inch television screen is between 5 and 7 years, we use the properties of the normal distribution where the mean (μ) is 6 years and the standard deviation (σ) is 0.5 years. We need to find the z-scores for 5 and 7 years and then use the standard normal distribution table or a calculator to find the probability that the service life falls between these two z-scores.

First, calculate the z-score for 5 years:
z = (X - μ) / σ
z = (5 - 6) / 0.5
z = -2.0

Next, calculate the z-score for 7 years:
z = (7 - 6) / 0.5
z = 2.0

Once we have the z-scores, we look up the corresponding probabilities in the normal distribution table or use a calculator with normal distribution functions. The probability of z being between -2.0 and 2.0 in a standard normal distribution is approximately 0.9545.

Therefore, the probability that a television will last between 5 and 7 years is approximately 95.45%.

Answer:

(a) 95% confidence interval for the population mean body temperature for healthy female is between a lower limit of 98.21 °F and an upper limit of 98.57 °F.

(b) The average is less than 98.6 °F

(c) Yes

Step-by-step explanation:

(a) Confidence interval = mean + or - Margin of Error (E)

mean = 98.39 °F

sd = 0.743 °F

n = 65

degree of freedom = n - 1 = 65 - 1 = 64

confidence level = 95%

t- value corresponding to 64 degrees of freedom and 95% confidence level is 1.9976.

E = t×sd/√n = 1.9976×0.743/√65 = 0.18 °F

Lower limit = mean - E = 98.39 - 0.18 = 98.21 °F

Upper limit = mean + E = 98.39 + 0.18 = 98.57 °F

95% confidence interval is between 98.21 °F and 98.57 °F.

(b) The average is less than 98.6 °F. The lower limit 98.21 °F and the upper limit 98.57 °F are both less than 98.6 °F

(c) It was valid to use the t-distribution approach to find the confidence Interval beci it gives a range of values for the population mean body temperature for healthy female.