In a little over a month, from June 5, 1879, to July 2, 1879, Albert Michelson measured the velocity of light in air 100 times (Stigler, Annals of Statistics, 1977). Today we know that the true value is 299,734.5 km/sec. Michelson’s data have a mean of 299,852.4 km/sec with a standard deviation of 79.01. (a) Find a two-sided 95% confidence interval for the true mean (the true value of the speed of light). (b) What does the confidence interval say about the accuracy of Michelson’s measurements?

the 95% confidence interval for the difference in proportions is approximately (0.1562, 0.2686).

To find the 95% confidence interval for the difference in proportions, we can use the formula:

[tex]\[\text{CI} = (\hat{p}_1 - \hat{p}_2) \pm z \times \sqrt{\frac{{\hat{p}_1(1 - \hat{p}_1)}}{n_1} + \frac{{\hat{p}_2(1 - \hat{p}_2)}}{n_2}}\][/tex]

Where:

- [tex]\(\hat{p}_1\) and \(\hat{p}_2\)[/tex] are the sample proportions.

- [tex]\(n_1\) and \(n_2\)[/tex] are the sample sizes.

- z is the z-score corresponding to the desired level of confidence.

Given:

- [tex]\(n_1 = 520\), \(n_2 = 460\)[/tex]

- [tex]\(\hat{p}_1 = \frac{318}{520}\), \(\hat{p}_2 = \frac{379}{460}\)[/tex]

- z = 1.96 for a 95% confidence interval

Let's plug in the values and calculate:

[tex]\[\hat{p}_1 = \frac{318}{520} \approx 0.6115\]\[\hat{p}_2 = \frac{379}{460} \approx 0.8239\]\[\text{CI} = (0.8239 - 0.6115) \pm 1.96 \times \sqrt{\frac{{0.6115 \times (1 - 0.6115)}}{520} + \frac{{0.8239 \times (1 - 0.8239)}}{460}}\]\[\text{CI} = (0.2124) \pm 1.96 \times \sqrt{\frac{{0.6115 \times 0.3885}}{520} + \frac{{0.8239 \times 0.1761}}{460}}\][/tex]

[tex]\[\text{CI} = (0.2124) \pm 1.96 \times \sqrt{0.000457 + 0.000368}\]\[\text{CI} = (0.2124) \pm 1.96 \times \sqrt{0.000825}\]\[\text{CI} = (0.2124) \pm 1.96 \times 0.0287\]\[\text{CI} = (0.2124) \pm 0.0562\][/tex]

Thus, the 95% confidence interval for the difference in proportions is approximately (0.1562, 0.2686).

Answer: Margin of error = 8.32, and Confidence interval using normal distribution is narrower than confidence interval using t-distribution.

Step-by-step explanation:

Since we have given that

n = 13

Mean repair cost = $85.00

Standard deviation = $15.30

At 95% confidence interval,

z= 1.96

Since it is normally distributed.

Margin of error is given by

[tex]z\times \dfrac{\sigma}{\sqrt{n}}\\\\=1.96\times \dfrac{15.30}{\sqrt{13}}\\\\=8.32[/tex]

95% confidence interval would be

[tex]\bar{x}\pm z\dfrac{\sigma}{\sqrt{n}}\\\\=85\pm 1.96\times \dfrac{15.30}{\sqrt{13}}\\\\=85\pm 8.32\\\\=(85-8.32,85+8.32)\\\\=(76.68,93.32)[/tex]

A 95​% confidence interval using the​ t-distribution was (75.8, 94.2 ).

Confidence interval using normal distribution is narrower than confidence interval using t-distribution.

Final answer:

A 95% confidence interval for the mean repair cost of microwave ovens with a known standard deviation is calculated using the z-score. The margin of error is found to be approximately $8.31, resulting in a confidence interval of ($76.69, $93.31).

Explanation:

To calculate the 95% confidence interval for the population mean when the population standard deviation is known, we can use the z-score associated with the 95% confidence level, which is 1.96. The formula for the margin of error (EBM) is EBM = z * (σ/√n), where σ is the population standard deviation, n is the sample size, and z is the z-score. Given that the sample standard deviation is $15.30, we assume it to be the population standard deviation because the question states that it is known.

With a sample mean (μ) of $85.00, a standard deviation of $15.30, and a sample size of 13, the margin of error is calculated as follows:

EBM = 1.96 * (15.30/√13) = 1.96 * 4.24 ≈ $8.31

The 95% confidence interval is therefore ($85.00 - $8.31, $85.00 + $8.31) = ($76.69, $93.31). The results using the z-distribution are similar to those obtained using the t-distribution, but usually, the t-distribution would be used when the sample size is small and the population standard deviation is unknown, which results in a wider interval due to the extra uncertainty.

Answer:

840

Step-by-step explanation:

Answer:

i know im late but it's 2 crates.

Step-by-step explanation:

To find the number of crates, divide the total number of cookies by the number of cookies in each box.

To find the number of crates that can be filled with the 3,258 cookies, we divide the total number of cookies by the number of cookies in each box. In this case, there are 12 cookies in each box.

So, we divide 3,258 by 12:

3,258 ÷ 12 = 271.5

Since we can't have half of a crate, we round down to the nearest whole number:

271.5 ≈ 271

Therefore, 271 crates can be filled from the total of 3,258 cookies.

Let [tex]\mu[/tex] denotes the average width of stripes .

As per given , we have

[tex]H_0:\mu=4\\H_a:\mu\neq4[/tex]

, since [tex]H_a[/tex] is two-tailed , so the test is a two-tailed test.

Also, population standard deviation is unknown , so we perform two-tailed t-test.

For Sample size : n= 45

Sample mean : [tex]\overline{x}=3.87[/tex]

Sample  standard deviation : s= 0.5 inches

Test statistic : [tex]t=\dfrac{\overline{x}-\mu}{\dfrac{s}{\sqrt{n}}}[/tex]

i.e.  [tex]t=\dfrac{3.87-4}{\dfrac{0.5}{\sqrt{45}}}\approx-1.74[/tex]

Significance level = [tex]\alpha=0.05[/tex]

By using t-value table,

Two-tailed critical t-value  = [tex]t_{\alpha/2,df}=t_{0.025,\ 44}=\pm2.0154[/tex]  [df = n-1]

Decision : Since the test statistic value (-1.74) lies with in the interval (-2.0154, 2.0154) , it means we are failed to reject the null hypothesis .

Conclusion: We have sufficient evidence to support the claim that the machine has slipped out of adjustment and the average width of stripes is no longer μ = 4 inches.

After calculating the z-score for the provided data, the result (-1.962) lies within the critical values for a 5% level of significance. Therefore, we cannot reject the null hypothesis, hence there's no sufficient evidence to state that the machine is out of adjustment.

To determine if the machine has slipped out of adjustment, we should conduct a hypothesis test. We can set the null hypothesis (H0) as μ = 4 (the machine is correctly adjusted), and the alternative hypothesis (Ha) as μ ≠ 4 (the machine has slipped). The sample size is large enough (>30) to use the z-score.
The z-score can be calculated using the formula: z = (x - μ)/(s/√n), where x is the sample mean, μ is the population mean, s is the standard deviation, and n is the sample size.
So, z = (3.87 - 4) / (0.5/√45) = -1.962. The critical z-value for a 5% level of significance (two-tailed test) is approximately +/- 1.96. Our calculated z-value falls within this range so we cannot reject the null hypothesis. Therefore, we don't have sufficient evidence to state that the machine has slipped out of adjustment.

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Answer:

Hip breadths less than or equal to 16.1 in. includes 90% of the males.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 14.5

Standard Deviation, σ = 1.2

We are given that the distribution of hip breadths is a bell shaped distribution that is a normal distribution.

Formula:

[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]

We have to find the value of x such that the probability is 0.10.

P(X > x)  

[tex]P( X > x) = P( z > \displaystyle\frac{x - 14.5}{1.2})=0.10[/tex]  

[tex]= 1 -P( z \leq \displaystyle\frac{x - 14.5}{1.2})=0.10 [/tex]  

[tex]=P( z \leq \displaystyle\frac{x - 14.5}{1.2})=0.90 [/tex]  

Calculation the value from standard normal z table, we have,  

[tex]P(z < 1.282) = 0.90[/tex]

[tex]\displaystyle\frac{x - 14.5}{1.2} = 1.282\\x = 16.0384 \approx 16.1[/tex]  

Hence, hip breadth of 16.1 in. separates the smallest 90​% from the largest 10%.

That is hip breaths greater than 16.1 in. lies in the larger 10%.

Answer:C

Step-by-step explanation:

Answer:

Hi! The answer is A, Blizzard.

Recently did this test on FLVS

Hope this helped!

Have a terrific Tuesday!!

~Lola

Answer:

Tornado

Step-by-step explanation:

Its not blizzard or hurricane.

Answer:

c) Is not a property (hence (d) is not either)

Step-by-step explanation:

Remember that the chi square distribution with k degrees of freedom has this formula

[tex]\chi_k^2 = \matchal{N}_1^2 +  \matchal{N}_2^2 + ... + \, \matchal{N}_{k-1}^2 +  \matchal{N}_k^2[/tex]

Where N₁ , N₂m .... [tex] N_k [/tex] are independent random variables with standard normal distribution. Since it is a sum of squares, then the chi square distribution cant take negative values, thus (c) is not true as property. Therefore, (d) cant be true either.

Since the chi square is a sum of squares of a symmetrical random variable, it is skewed to the right (values with big absolute value, either positive or negative, will represent a big weight for the graph that is not compensated with values near 0). This shows that (a) is true

The more degrees of freedom the chi square has, the less skewed to the right it is, up to the point of being almost symmetrical for high values of k. In fact, the Central Limit Theorem states that a chi sqare with n degrees of freedom, with n big, will have a distribution approximate to a Normal distribution, therefore, it is not very skewed for high values of n. As a conclusion, the shape of the distribution changes when the degrees of freedom increase, because the distribution is more symmetrical the higher the degrees of freedom are. Thus, (b) is true.

Answer:  The answers are given below.

Step-by-step explanation:  Given that a  3 × 3 matrix P satisfies the matrix equation P² = P.

We are to

(a) find the possibilities for the determinant of P.

(b) explain the reason behind there are no other possibilities.

(c) give an example of P, for each possible determinant.

(a) According to the given information, we have

[tex]P^2=P\\\\\Rightarrow P^2-P=0\\\\\Rightarrow P(P-I)=0\\\\\Rightarrow P=0,~~~P=I.[/tex]

So, P can be either a zero matrix of order 3 or an identity matrix of order 3.

If P = 0, then det(P) = 0     and     if P = I, then det(P) = 1.

Therefore the possible determinants of P are 0 and 1.

(b) There can be any other determinant other than 0 and 1, because if so, then the given equation P² = P will not be satisfied.

(c) If |P| = 0, then the matrix P can be can be of the form as follows:

[tex]P=\left[\begin{array}{ccc}0&0&0\\0&0&0\\0&0&0\end{array}\right] .[/tex]

If |P| = 1, the the matrix P can be of the form as follows :

[tex]P=\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right] .[/tex]

Thus, all the parts are answered.

Answer:

We reject the null hypothesis that the population means are equal and accept the alternative hypothesis that the population means are different.

Step-by-step explanation:

We have large sample sizes [tex]n_{1} = 49[/tex] and [tex]n_{2} = 64[/tex], the unbiased point estimate for [tex]\mu_{1}-\mu_{2}[/tex] is [tex]\bar{x}_{1} - \bar{x}_{2}[/tex], i.e., 12-14 = -2.

The standard error is given by [tex]\sqrt{\frac{\sigma^{2}_{1}}{n_{1}}+\frac{\sigma^{2}_{2}}{n_{2}}}[/tex], i.e.,

[tex]\sqrt{\frac{(3)^{2}}{49}+\frac{(4)^{2}}{64}}[/tex] = 0.6585.

We want to test [tex]H_{0}: \mu_{1}-\mu_{2} = 0[/tex] vs [tex]H_{1}: \mu_{1}-\mu_{2} \neq 0[/tex] (two-tailed alternative). The rejection region is given by RR = {z | z < -2.5758 or z > 2.5758} where -2.5758 and 2.5758 are the 0.5th and 99.5th quantiles of the standard normal distribution respectively. The test statistic is [tex]Z = \frac{\bar{x}_{1} - \bar{x}_{2}-0}{\sqrt{\frac{\sigma^{2}_{1}}{n_{1}}+\frac{\sigma^{2}_{2}}{n_{2}}}}[/tex] and the observed value is [tex]z_{0} = \frac{-2}{0.6585} = -3.0372[/tex]. Because -3.0372  fall inside RR, we reject the null hypothesis.

The test statistic follow a standard normal distribution because we are dealing with large sample sizes.

In this scenario of comparing two independent samples and given that the sample sizes are large, the sample test statistic follows the Standard Normal distribution or Z-distribution. The Z-test statistic representing the difference in sample means (in units of standard error) is compared with critical values for a two-tailed test at 0.01 significance level to determine if there's sufficient evidence to reject the null hypothesis that the two population means are equal.

The test in your question pertains to a hypothesis testing scenario featuring two independent samples. This scenario typically involves two population means given that population standard deviations are known. The distribution followed by the sample test statistic in such cases is the Standard Normal distribution or Z-distribution, as the sample sizes (n1 = 49, n2 = 64) are sufficiently large. To test the claim that population means are different (at a significance level of 0.01), you'd typically construct a Z-test statistic that represents the difference in sample means (x1 - x2) in units of its standard error. The Z-test statistic is calculated as follows:  

Here, x1 and x2 are the sample means, σ1 and σ2 are the population standard deviations and n1 and n2 are the samples sizes. The resulting Z-score can be compared with critical Z-scores for a two-tailed test at the given level of significance (0.01) to determine whether or not the null hypothesis (two population means are equal) can be rejected.

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Answer:

The proof is given below.

Step-by-step explanation:

this theorem says that, if a point is on the angular bisector of an angle, then it is equidistant from the sides of the angle.

(the proof of  angle bisector theorem can be explained, but it is difficult to type the whole thing. so watch this video for the proof of this theorem :  https://youtu.be/6GS4lS4btNI    )

To calculate the total cost of drilling a 500-meter well, add the fixed cost to the sum of the marginal costs for every meter drilled. The total cost comes out to 3,500,000 riyals.

The total cost of drilling a 500-meter well comprises both fixed costs and the marginal cost per meter of depth. We are given that the fixed costs amount to 1,000,000 riyals. The marginal cost function is given as C'(x) = 4000 + 10x, which means the cost per additional meter drilled increases linearly with depth.

To find the total cost of drilling a 500-meter well, we need to compute the integral (i.e., the area under the curve) of the marginal cost function from 0 to 500 and add the fixed costs. This calculation represents the sum of the increasing cost per meter for every meter drilled.

By evaluating the integral ∫ (4000 + 10x) dx from 0 to 500, we get 2,500,000 riyals. This is the total variable cost of drilling a 500m well. Adding it to the fixed cost (1,000,000 riyals), the grand total comes out to be 3,500,000 riyals.

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Answer:

The probability the average mpg is less than 28 mpg is 0.8413.

Step-by-step explanation:

Given : The manufacturer of a new compact car claims the miles per gallon (mpg) for the gasoline consumption is approximately normal with

Mean [tex]\mu=26[/tex] mpg and standard deviation [tex]\sigma=12[/tex] mpg.

Number of sample n=36

To find : What is the probability the average mpg is less than 28 mpg?

Solution :

Applying z-score formula,

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

The portability is given by, [tex]P(X<28)[/tex]

[tex]=P(\dfrac{x-\mu}{\frac{\sigma}{\sqrt{n}}}<\dfrac{28-26}{\frac{12}{\sqrt{36}}})[/tex]

[tex]=P(\dfrac{x-\mu}{\frac{\sigma}{\sqrt{n}}}<\dfrac{2}{\frac{12}{6}})[/tex]

[tex]=P(z<\dfrac{2}{2})[/tex]

[tex]=P(z<1)[/tex]

Using z-table,

[tex]=0.8413[/tex]

Therefore, the probability the average mpg is less than 28 mpg is 0.8413.

Answer:

he probability the average mpg is less than 28 mpg is 0.8413.

Step-by-step explanation:

Given : The manufacturer of a new compact car claims the miles per gallon (mpg) for the gasoline consumption is approximately normal with

Mean  mpg and standard deviation  mpg.

Number of sample n=36

To find : What is the probability the average mpg is less than 28 mpg?

Solution :

Applying z-score formula,

The portability is given by,  

Using z-table,

Therefore, the probability the average mpg is less than 28 mpg is 0.8413.

Step-by-step explanation:

Answer:

2/9

Step-by-step explanation:

The Poisson’s distribution is a discrete probability distribution. A discrete probability distribution means that the events occur with a constant mean rate and independently of each other. It is used to signify the chance (probability) of a given number of events occurring in a fixed interval of time or space.

In the long run, fraction of time that it rains = E(Number of days in rainy spell) / {E(Number of days in a rainy spell) + E(Number of days in a dry spell)}

E(Number of days in rainy spell) = 2

E(Number of days in a dry spell) = 7

In the long run, fraction of time that it rains = 2/(2 + 7) = 2/9

Given the parameters of the rainy spell and dry spell, the long-run fraction of time that it rains can be calculated by dividing the mean of the rainy days by the sum of the average rainy and dry days. Hence, it rains roughly 22.22% of the time in the long-term.

The question is asking about the long-run fraction of time that it rains, based on a rainy spell following a Poisson distribution with a mean of 2 days, and a dry spell following a geometric distribution with an average of 7 days, with the sequences being independent.

We are being asked to calculate the proportion of time that it rains in the long-run, given these distribution parameters. The Poisson and geometric distributions are often used in this type of probability assessment.

To tackle this, we need to divide the mean of the rainy days by the sum of the average rainy and dry days. Thus, the long-run fraction of time it rains is given by [tex]2/(2+7) = 2/9.[/tex]

So, in the long run, it rains roughly 22.22% (or 2/9) of the time.

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Answer:

We accept the alternate hypothesis. We conclude that the mean lifetime of tires is  is less than 33,000 miles.

Step-by-step explanation:

We are given the following in the question:  

Population mean, μ = 33,000 miles

Sample mean, [tex]\bar{x}[/tex] =  32, 450 miles

Sample size, n = 18

Alpha, α = 0.05

Sample standard deviation, s = 1200 miles

a) First, we design the null and the alternate hypothesis

[tex]H_{0}: \mu = 33000\text{ miles}\\H_A: \mu < 33000\text{ miles}[/tex]

b) Level of significance:

[tex]\alpha = 0.05[/tex]

c) We use One-tailed t test to perform this hypothesis.

d) Formula:

[tex]t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }[/tex] Putting all the values, we have

[tex]t_{stat} = \displaystyle\frac{32450 - 33000}{\frac{1200}{\sqrt{18}} } = -1.9445[/tex]

Now, [tex]t_{critical} \text{ at 0.05 level of significance, 17 degree of freedom } = -1.7396[/tex]

Rejection area:

[tex]t < -1.7396[/tex]

Since,                  

[tex]t_{stat} < t_{critical}[/tex]

e) We fail to accept the null hypothesis and reject it as the calculated value of t lies in the rejection area.

f) We accept the alternate hypothesis. We conclude that the mean lifetime of tires is  is less than 33,000 miles.

Answer:

p-value:  0.6527

Step-by-step explanation:

Hello!

You have two samples to study, from each sample the weight of each child was measured and counted the total of overweight kids (x: "success") in each group:

Sample 1 (Boys aged 6-11)

n₁= 546

x₁= 89

^p₁= x₁/n₁ = 89/546 ≅0.16

Sample 2 (girls aged 6-11)

n=508

x= 74

^p= x/n = 74/508 ≅ 0.15

If the hypothesis statement is "The proportion of boys that are overweight differs from the proportion of girls that are overweight", the test hypothesis is:

H₀: ρ₁ = ρ₂

H₁: ρ₁ ≠ ρ₂

This type of hypothesis leads to a two-tailed rejection region, then the p-value will also be two-tailed. To calculate the p-value you have to first calculate the value of the statistic under the null hypothesis, in this case, is a test for the difference between two proportions:

Z=      (^ρ₁ - ^ρ₂) - (ρ₁ - ρ₂)         ≈ N(0;1)

    √(ρ` * (1 - ρ`) * (1/n₁ + 1/n₂))

ρ`= x₁ + x₂   =  89+74     = 0.154 ≅ 0.15

     n₁ + n₂     546 + 508

Z⁰ᵇ =          (0.16-0.15) - (0)                    

       √(0.15 * (1 - 0.15) * (1/546 + 1/508))

Z⁰ᵇ = 0.45

I've mentioned before that in this test you have a two-tailed p-value. The value calculated (0.45) corresponds to the right or positive tail and the left tail is symmetrical to it concerning the distribution mean, in this case, is 0, so it is -0.45. To obtain the p-value you need to calculate the probability of both values and add them:

P(Z>0.45) + P(Z<-0.45) = (1- P(Z<0.45)) + P(Z<-0.45) = (1-0.67364) + 0.32636 = 0.65272 ≅ 0.6527

p-value:  0.6527

Since there is no signification level in the problem, I'll use the most common to reach a decision. α: 0.05

Since the p-value is greater than α, you do not reject the null Hypothesis, in other words, there is no significative difference between the proportion of overweight boys and the proportion of overweight girls.

I hope it helps!

Answer:

0.0004498

Step-by-step explanation:

Let us define the events:

A = The test returns positive.

B = The accused was present.

Since everyone in the town has an equal probability of being at the murder scene, and the town has a population of 10,000

P(B) = 1/10000 = 0.0001

We have that the probability the test returning positive given that the accused was actually present at the scene is 0.9

P(A | B) = 0.9

and the probability that the test returns negative given that the accused was not present at the scene is 0.8

[tex]\large P(A^c|B^c)=0.8[/tex]

where

[tex]\large  A^c,\;B^c[/tex] are the complements of A and B respectively.

We want to determine the probability that the DNA test returned a positive result given that the accused was at the murder scene, that is, P(B | A).

We know that P(A | B) = 0.9, so

[tex]\large \frac{P(A\cap B)}{P(B)}=0.9\Rightarrow P(A\cap B)=0.9P(B)=0.9*0.0001\Rightarrow\\\\P(A\cap B)=0.00009[/tex]

Now, we have

[tex]\large P(B|A)=\frac{P(A\cap B)}{P(A)}=\frac{0.00009}{P(A)}[/tex]

So if we can determine P(A), the result will follow.

By De Morgan's Law

[tex]\large A^c\cap B^c=(A\cup B)^c[/tex]

so

[tex]\large 0.8=P(A^c|B^c)= \frac{P(A^c\cap B^c)}{P(B^c)}=\frac{P((A\cup B)^c)}{P(B^c)}=\frac{1-P(A\cup B)}{1-P(B)}\Rightarrow\\\\\frac{1-P(A\cup B)}{1-0.0001}=0.8\Rightarrow P(A\cup B)=1-0.8(1-0.0001)\Rightarrow\\\\P(A\cup B)=0.20008[/tex]

Using the formula

[tex]\large P(A\cup B)=P(A)+P(B)-P(A\cap  B)[/tex]

and replacing the values we have found

[tex]\large 0.20008=P(A)+0.0001-0.00009\Rightarrow\\\\P(A)=0.20007[/tex]

and finally, the desired result is

[tex]\large P(B|A)=\frac{0.00009}{P(A)}=\frac{0.00009}{0.20007}\Rightarrow\\\\\boxed{P(B|A)=0.0004498}[/tex]

Answer:

a) [tex]\hat \mu = \bar x =\sum_{i=1}^{10} \frac{x_i}{10}=120.2[/tex]

b) [tex]\hat \tau =n\hat \mu =25000x120.2=3005000[/tex]

c) [tex]\hat p= \hat \theta =\frac{8}{10}=0.8[/tex]

d) Median =120

Step-by-step explanation:

1) Some important concepts

The mean refers to the "average that is used to derive the central tendency of data analyzed. It is determined by adding all the data points in a population and then dividing the total by the number of points".

Method of moments "involves equating sample moments with theoretical moments". For example the first sample moment about the origin is defined as [tex]M_1=\frac{1}{n} \sum_{i=1}^{n}x_i =\bar X [/tex]

The median is "the middle number in a sorted, ascending or descending, list of numbers and can be more descriptive of that data set than the average".

When we are trying to estimate the population proportion, p.

All estimation is based on the fact that the normal can be used to approximate the binomial distribution when np and nq are both at least 5. Where p is the probability of success and q the probability of failure.

2) Part a

Using the method of the moments a point of estimate for the [tex]\mu[/tex] is:

[tex]\hat \mu = \bar x =\sum_{i=1}^{10} \frac{x_i}{10}=120.2[/tex]

3) Part b

If [tex]\hat \mu[/tex] is an individual estimate for the average gas usage during January and [tex]\tau[/tex] represent "the total amount of gas used by all of these houses during January" then the estimation for the total would be given by:

[tex]\hat \tau =n\hat \mu =25000x120.2=3005000[/tex]

3) Part c

For this part we want to estimate p ="the proportion of all houses that used at least 100 therms". If X is the random variable who represent the number of houses that exceed the usage of 100, we see that 8 out of 10 values are above 100, so the random variable X would be distributed binomial

[tex]X \sim Bin(10,0.8)[/tex] where n=10 and

[tex]\hat p= \hat \theta =\frac{8}{10}=0.8[/tex]

4) Part d

In order to find the median we need to put the data in order first, like this:

86,99,103,109,118,122,125,138,149,153

Since we have 10 observations and this number is even the procedure that we need to use in order to find the median is:

a) Find the value at position[n/2]=[10/2]=[5] on the data set ordered. For this case the value at position [5] is 118

b) Find the value at position[n/2 +1]=[10/2 +1]=[6] on the data set ordered. For this case the value at position [6] is 122

c) Find the average from the values obtained on steps a) and b). for this case (118+122)/2=120

So the Median = 120

The mean of the data set is 120.2

The total gas used =  3005000

The proportion of at least therms = 0.8

The median of the set = 120

a. To get the average gas usage, we are asked to calculate the mean of the observation.

Average

[tex]\frac{ 153+103+125+149+118+109+86+122+138+99}{10} \\\\[/tex]

= 120.2

b. The question says that 25000 houses make use of natural gas, then

total gases used by these houses =

120.2 * 25000

= 3005000

c. The proportion of the houses that used above 100 therms,

The house above 100 here are, 125, 149, 118, 109, 122, 138

They are 8 in number.

8/10 = 0.8

0.8 is the proportion that used at least 100 therms.

d. We have to find the median for the set here. We arrange the details in ascending order.

86,99,103,109,118,122,125,138,149,153

Median = 118+122/2

= 240/2

= 120

Answer:

We have the matrix [tex]A=\left[\begin{array}{ccc}-4&-4&-4\\0&-8&-4\\0&8&4\end{array}\right][/tex]

To find the eigenvalues of A we need find the zeros of the polynomial characteristic [tex]p(\lambda)=det(A-\lambda I_3)[/tex]

Then

[tex]p(\lambda)=det(\left[\begin{array}{ccc}-4-\lambda&-4&-4\\0&-8-\lambda&-4\\0&8&4-\lambda\end{array}\right] )\\=(-4-\lambda)det(\left[\begin{array}{cc}-8-\lambda&-4\\8&4-\lambda\end{array}\right] )\\=(-4-\lambda)((-8-\lambda)(4-\lambda)+32)\\=-\lambda^3-8\lambda^2-16\lambda[/tex]

Now, we fin the zeros of [tex]p(\lambda)[/tex].

[tex]p(\lambda)=-\lambda^3-8\lambda^2-16\lambda=0\\\lambda(-\lambda^2-8\lambda-16)=0\\\lambda_{1}=0\; o \; \lambda_{2,3}=\frac{8\pm\sqrt{8^2-4(-1)(-16)}}{-2}=\frac{8}{-2}=-4[/tex]

Then, the eigenvalues of A are [tex]\lambda_{1}=0[/tex] of multiplicity 1 and [tex]\lambda{2}=-4[/tex] of multiplicity 2.

Let's find the eigenspaces of A. For [tex]\lambda_{1}=0[/tex]: [tex]E_0 = Null(A- 0I_3)=Null(A)[/tex].Then, we use row operations to find the echelon form of the matrix

[tex]A=\left[\begin{array}{ccc}-4&-4&-4\\0&-8&-4\\0&8&4\end{array}\right]\rightarrow\left[\begin{array}{ccc}-4&-4&-4\\0&-8&-4\\0&0&0\end{array}\right][/tex]

We use backward substitution and we obtain

1.

[tex]-8y-4z=0\\y=\frac{-1}{2}z[/tex]

2.

[tex]-4x-4y-4z=0\\-4x-4(\frac{-1}{2}z)-4z=0\\x=\frac{-1}{2}z[/tex]

Therefore,

[tex]E_0=\{(x,y,z): (x,y,z)=(-\frac{1}{2}t,-\frac{1}{2}t,t)\}=gen((-\frac{1}{2},-\frac{1}{2},1))[/tex]

For [tex]\lambda_{2}=-4[/tex]: [tex]E_{-4} = Null(A- (-4)I_3)=Null(A+4I_3)[/tex].Then, we use row operations to find the echelon form of the matrix

[tex]A+4I_3=\left[\begin{array}{ccc}0&-4&-4\\0&-4&-4\\0&8&8\end{array}\right] \rightarrow\left[\begin{array}{ccc}0&-4&-4\\0&0&0\\0&0&0\end{array}\right][/tex]

We use backward substitution and we obtain

1.

[tex]-4y-4z=0\\y=-z[/tex]

Then,

[tex]E_{-4}=\{(x,y,z): (x,y,z)=(x,z,z)\}=gen((1,0,0),(0,1,1))[/tex]

Answer:

b) The annual rainfall in a city

Step-by-step explanation:

Remember, a discrete variable is one that can only take a finite number of values between any two values of a characteristic and a continuous variable is one that can take an infinite number of values between any two values of a characteristic.

a) Observe that the variable x='classes taken in one semester' can take the values 0,1,2,...,n.

Then the variable x is discrete

b) Observe that the variable x='annual rainfall in a city' can take the values 2in, 1.6in, 5.1 in, 0.1in

Then, the variable x can be take a infinite number of values between two number. So x isn't a discrete variable.

c) The variable x='attendance at a football game' can take the values 3000,5000... n. And never will be a decimal number because There cannot be a personal decimal number. Therefore, x is a discrete variable.

d) The variable x='patients treated at an emergency room in a day' can take the values 1,2,3,...,n. And never will be a decimal number because There cannot be a personal decimal number. Therefore, x is a discrete variable.

Answer: (6.304, 6.696)  

Step-by-step explanation:

The confidence interval for population mean is given by :-

[tex]\overline{x}\pm z^*\dfrac{\sigma}{\sqrt{n}}[/tex]

, where [tex]\sigma[/tex] = Population standard deviation.

n= sample size

[tex]\overline{x}[/tex] = Sample mean

z* = Critical z-value .

Let x denotes the number of  hours slept by UCF students.

Given :  [tex]\sigma=2\ hours[/tex]

n= 400

[tex]\overline{x}= 6.5\ hours[/tex]

Two-tailed critical value for 95% confidence interval = [tex]z^*=1.96[/tex]

Then, the 95%confidence interval for the true number of hours slept by UCF students will be :-

[tex]6.5\pm(1.96)\dfrac{2}{\sqrt{400}}\\\\=6.5\pm(1.96)\dfrac{2}{20}\\\\=6.5\pm0.196=(6.5-0.196,\ 6.5+0.196)=(6.304,\ 6.696)[/tex]

Hence, the 95% confidence interval for the true number of hours slept by UCF students : (6.304, 6.696)  

The equation modeling the displacement d as a function of time t for an object in simple harmonic motion with a period of 7 seconds and amplitude of 3cm is d = -3cos(2pi/7 * t + pi).

The displacement d of an object moving in simple harmonic motion can be modeled by the equation d=Acos(wt).

Given that the period T is 7 seconds, we can use the formula T=2pi/w to solve for the angular frequency w. Rearranging the equation, we have w = 2pi/T. Plugging in the given period T=7, we get w = 2pi/7.

Since the object initially moves in a negative direction, we would have a phase shift of pi in the cosine function. Therefore, the equation modeling the displacement d as a function of time t is d = -3cos(2pi/7 * t + pi).

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The circulation of [tex]\( \vec{F} \)[/tex] around [tex]\( C \)[/tex]  is[tex]\( 3\pi \)[/tex], computed using Green's theorem by transforming the line integral into a double integral over the enclosed region.

To calculate the circulation of [tex]\( \vec{F} \)[/tex] around C, we first need to compute the line integral of [tex]\( \vec{F} \)[/tex] along C. Using Green's theorem, we can rewrite this line integral as a double integral over the region enclosed by C .

Given [tex]\( \vec{F}(x, y) = (3x - 4y)\vec{i} + 2x\vec{j} \)[/tex], we can compute the partial derivatives [tex]\( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \)[/tex], where[tex]\( P = 3x - 4y \) and \( Q = 2x \).[/tex]

[tex]\[\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 2 - (-4) = 6\][/tex]

Since the region enclosed by C is the counter-clockwise oriented sector of a circle with radius 2 and central angle [tex]\( \frac{\pi}{4} \)[/tex], we can parameterize the curve [tex]\( C \)[/tex] as [tex]\( x(t) = 2\cos(t) \) and \( y(t) = 2\sin(t) \)[/tex] where [tex]\( 0 \leq t \leq \frac{\pi}{4} \).[/tex]

Now, applying Green's theorem, we have:

[tex]\[\text{Circulation} = \iint_D (6) \, dA\][/tex]

Using polar coordinates, the double integral becomes:

[tex]\[\begin{aligned}\text{Circulation} &= \int_{0}^{\frac{\pi}{4}} \int_{0}^{2} (6) \cdot r \, dr \, dt \\&= \int_{0}^{\frac{\pi}{4}} \left[3r^2\right]_{0}^{2} \, dt \\&= \int_{0}^{\frac{\pi}{4}} 3(2)^2 \, dt \\&= \int_{0}^{\frac{\pi}{4}} 12 \, dt \\&= [12t]_{0}^{\frac{\pi}{4}} \\&= 12\left(\frac{\pi}{4}\right) - 12(0) \\&= 3\pi\end{aligned}\][/tex]

Therefore, the circulation of [tex]\( \vec{F} \)[/tex] around [tex]\( C \)[/tex] is [tex]\( 3\pi \).[/tex]

The question probable maybe:

Given in the attachment

Answer:

22.23

Step-by-step explanation:

Given that in a binomial experiment study, in a certain community showed that 5% of the people suffer from insomnia

i.e p = 0.05

q=0.95

n=10400

[tex]np=10400*0.05=202\\npq =494[/tex]

Var(x)=[tex]npq=494\\std dev=\sqrt{494} \\=22.23[/tex]

the standard deviation of the number of people who suffer from insomnia

=22.23

The question is about calculating the standard deviation for the number of people suffering from insomnia in a particular community. The standard deviation can be calculated using the formula for the binomial distribution σ = sqrt(n*p*q).

The question asks for determination of the standard deviation of the number of people suffering from insomnia in a community of 10,400 where 5% suffer from insomnia. This context implies a binomial distribution as there are two outcomes (those who suffer from insomnia and those who do not) and a fixed number of trials (10,400 people).

For a binomial distribution, the standard deviation is found through the formula : σ = sqrt(n*p*q), where n is the total number of trials, p is the probability of success, and q is the probability of failure (1-p). Here, n=10,400, p=0.05 and q=0.95.

Here the standard deviation would be calculated as σ = sqrt{10,400 * 0.05 * 0.95}.

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Answer:

The answer is B.8.  

Step-by-step explanation:

This is the answer because 8 is the greatest factor that will go into both 48 and 56. 8x6=48, 8x7=56.

The greatest common factor (GCF) of 48 and 56 is 8. It is the highest number that divides both numbers without leaving a remainder. Thus, the correct answer is Option B. 8.

The greatest common factor (GCF) of two numbers is the largest number that divides both of them without leaving a remainder. To find the GCF of 48 and 56, follow these steps:

Therefore, the correct answer is Option B. 8. This makes 8 the highest number that can evenly divide both 48 and 56.

Answer:  The length of the loner side is 7 in. and the length of the shorter side is 6 in.

Step-by-step explanation:  Given that a graphics designer is designing an advertising brochure for an art show. Each page of the brochure is rectangular with an area of 42 in squared and a perimeter of 26 in.

We are to find the dimensions of the brochure.

Let l and b represents the lengths of the longer side and shorter side respectively of each page of the brochure.

Then, according to the given information, we have

[tex]l\times b=42~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)[/tex]

and

[tex]2(l+b)=26\\\\\Rightarrow l+b=13\\\\\Rightarrow l=13-b~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(ii)[/tex]

Substituting the value of l from equation (ii) in equation (i), we get

[tex](13-b)b=42\\\\\Rightarrow b^2-13b+42=0\\\\\Rightarrow b^2-6b-7b+42=0\\\\\Rightarrow (b-6)(b-7)=0\\\\\Rightarrow b-6=0,~~~b-7=0\\\\\Rightarrow b=6,7.[/tex]

Since b is the length of the shorter side, so b = 6 in.

From equation (ii), we get

[tex]l=13-6=7.[/tex]

Thus, the length of the loner side is 7 in. and the length of the shorter side is 6 in.

To find the dimensions of the brochure, we use the formulas for the area and perimeter of a rectangle. Solving the system of equations produced by these formulas, we find that the length of the brochure is 7 inches and the width is 6 inches.

In order to find the dimensions of the brochure, we can use the formulas for the area and perimeter of a rectangle. Given area, A = 42 inches squared and perimeter, P = 26 inches. The formulas for the area and perimeter of the rectangle are A = length x width and P = 2(length + width).

Let's denote the length of the rectangle as 'l' and the width as 'w'. Now we know that:

l x w = 42 inches (according to area formula)

2(l + w) = 26 inches (according to perimeter formula)

This is a system of two equations which can be solved simultaneously. After solving these equations, we find that the length (longer side) is 7 inches, and the width (shorter side) is 6 inches.

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Answer: x = 0

y = 2

z = -1

Step-by-step explanation:

The system of equations are

x+y+z=1 - - - - - - - - - - 1

-2x+4y+6z=2 - - - - - - - - - 2

-x+3y-5z=11 - - - - - - - - - 3

Step 1

We would eliminate x by adding equation 1 to equation 3. It becomes

4y -4z = 12 - - - - - - - - - 4

Step 2

We would multiply equation 1 by 2. It becomes

2x + 2y + 2z = 2 - - - - - - - - - 5

We would add equation 2 and equation 5. It becomes

6y + 8z = 4 - - - - - - - - - 6

Step 3

We would multiply equation 4 by 6 and equation 6 by 4. It becomes

24y - 24z = 72 - - - - - - - - 7

24y + 32z = 16 - - - - - - - - 8

We would subtract equation 8 from equation 7. It becomes

-56z = 56

z = -56/56 = -1

Substituting z = -1 into 7, it becomes

24y - 24×-1 = 72

24y + 24 = 72

24y = 72 - 24 = 48

y = 48/24 = 2

Substituting y = 2 and z = -1 into equation 1, it becomes

x + 2 - 1 = 1

x = 1 - 1 = 0

The suggested mileage interval, excluding no more than 10% of the mileage on tires, is approximately 18,420 to 31,580 miles.

To determine the mileage interval that excludes no more than 10% of the mileage on tires, we can use the standard normal distribution and the properties of the normal curve. The mileage data is normally distributed with a mean [tex](\(\mu\))[/tex] of 25,000 miles and a standard deviation [tex](\(\sigma\))[/tex] of 4,000 miles.

To find the interval, we can use the Z-score formula:

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

To exclude no more than 10% of the mileage, we need to find the Z-score corresponding to the 5th percentile on each side of the mean, as 10% is split between the lower and upper tails of the distribution.

Using a standard normal distribution table or a calculator, the Z-score for the 5th percentile is approximately -1.645 (negative due to being in the lower tail).

Now, we can use the Z-score formula to find the values of [tex]\(X\)[/tex] (mileage) corresponding to these Z-scores:

[tex]\[ X_{\text{lower}} = \mu + Z \times \sigma \][/tex]

[tex]\[ X_{\text{upper}} = \mu - Z \times \sigma \][/tex]

Substitute the values:

[tex]\[ X_{\text{lower}} = 25,000 - (-1.645) \times 4,000 \][/tex]

[tex]\[ X_{\text{upper}} = 25,000 + (-1.645) \times 4,000 \][/tex]

Calculating these values:

[tex]\[ X_{\text{lower}} \approx 31,580 \][/tex]

[tex]\[ X_{\text{upper}} \approx 18,420 \][/tex]

Therefore, the suggested mileage interval is approximately 18,420 miles to 31,580 miles to exclude no more than 10% of the mileage on the tires.

Answer:

False

Step-by-step explanation:

Confidence intervals provide a range for a population parameter at a given significance level. The parameter can be mean, standard deviation etc.

In this example population is the prices of the rents of all the unfurnished one-bedroom apartments in the Boston area

significance level is 95%. Thus, the chance being the true population parameter in the given interval is 95%.

But, "This interval describes the price of 95% of the rents of all the unfurnished one-bedroom apartments in the Boston area." statement is false because the population parameter is missing. Confidence interval may describe population mean for example but it does not describe the whole population.