A bettor with utility function U(x) = ln(x), where x is total wealth, has a choice between the following two alternatives:
A. Win $ 10,000 with probability 0.2
Win $1000 with probability 0.8 B
B. Win $3000 with probability 0.9
Lose $2000 with probability 0.1
1. If the bettor currently has $2500, should he choose A or B?
2. Repeat a, assuming the bettor has $5000.
3. Repeat a, assuming the bettor has $10,000.
4. Do you think that this pattern of choices between A and B is reasonable? Why or why not?

Answer:

[tex](0.2935, 0.4265)[/tex]

Step-by-step explanation:

Given that a random sample of 200 books purchased at a local bookstore showed that 72 of the books were murder mysteries.

Sample proportion p follows the following characteristics

p 0.36

q 0.64

pq/n 0.001152

Std dev 0.033941125

For 95% confidence level we have critical value as

Z critical = 1.96

Hence margin of error = 1.96*std dev

Confidence interval = (sample proportion - margin of error, sample proportion + margin of error)

=[tex](0.2935, 0.4265)[/tex]

To construct the 95% confidence interval for the proportion of murder mystery books sold at a bookstore, we use the sample proportion, Z-score for 95% confidence, and sample size to find the margin of error and then add and subtract this from the sample proportion. The 95% confidence interval is approximately (0.2935, 0.4265).

To construct a 95% confidence interval for the true proportion of books sold by the store that are murder mysteries, we can use the formula for a confidence interval for a proportion, which is [tex]p \pm Z*(\sqrt{\frac{p(1-p)}{n}})[/tex]

Sample proportion (p) = 72÷200 = 0.36

Sample size (n) = 200

For a 95% confidence interval, we use the Z-score corresponding to 95% confidence level from the standard normal distribution table, which is approximately 1.96.

The margin of error (E) is calculated as follows:

[tex]E = 1.96 * \sqrt{\frac{0.36(1-0.36)}{200}}\\E = 1.96 * \sqrt{0.001152}[/tex]

E = 1.96 * 0.033941

E ≈ 0.0665

Now we can construct the confidence interval:

The lower boundary is p - E = 0.36 - 0.0665 ≈ 0.2935

The upper boundary is p + E = 0.36 + 0.0665 ≈ 0.4265

Therefore, the 95% confidence interval is (0.2935, 0.4265).

Answer:

The correct option is b) 1.70

Step-by-step explanation:

Consider the provided information.

The weekly sulfur dioxide emissions follow a normal distribution with a mean of 1000 ppm (parts per million) and a standard deviation of 25.

Thus, μ=1000 and σ = 25

The CEO wants to know if the mean level of emissions is different from 1000.

Therefore the null and alternative hypothesis is:

[tex]H_0:\mu =1000[/tex] and [tex]H_a:\mu \neq1000[/tex]

The sample size is n = 50 and [tex]\bar x=1006[/tex] ppm.

Now calculate the test statistic by using the formula: [tex]z=\frac{\bar x-\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

Substitute the respective values in the above formula.

[tex]z=\frac{1006-1000}{\frac{25}{\sqrt{50}}}[/tex]

[tex]z=\frac{6}{\frac{25}{\sqrt{50}}}=1.697\approx 1.70[/tex]

Hence, the correct option is b) 1.70

Answer:

option (c) n = 201

Step-by-step explanation:

Data provided in the question:

Standard deviation, s = 5.5 ounce

Confidence level = 99%

Length of confidence interval = 2 ounces

Therefore,

margin of error, E = (Length of confidence interval ) ÷ 2

= 2 ÷ 2

= 1 ounce

Now,

E = [tex]\frac{zs}{\sqrt n}[/tex]

here,

z = 2.58 for 99% confidence interval

n = sample size

thus,

1 = [tex]\frac{2.58\times5.5}{\sqrt n}[/tex]

or

n = (2.58 × 5.5)²

or

n = 201.3561 ≈ 201

Hence,

option (c) n = 201

Answer:

C. n=423

Step-by-step explanation:

1) Notation and important concepts

Margin of error for a proportion is defined as "percentage points your results will differ from the real population value"

Confidence=90%=0.9

[tex]\alpha=1-0.9=0.1[/tex] represent the significance level defined as "a measure of the strength of the evidence that must be present in your sample before you will reject the null hypothesis and conclude that the effect is statistically significant".

[tex]\hat p=0.5[/tex] represent the sample proportion of consumers in the Oconee County area who would react favorably to a marketing campaig. For this case since we don't have enough info we use the value of 0.5 since is equiprobable the event analyzed.

[tex]z_{\alpha/2}[/tex] represent a quantile of the normal standard distribution that accumulates [tex]{\alpha/2}[/tex] on each tail of the distribution.

2) Formulas and solution for the problem

For this case the margin of error for a proportion is given by this formula

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

For this case the confidence level is 90% or 0.9 so then the significance would be

[tex]\alpha=1-0.9=0.1[/tex] and [tex]\alpha/2=0.05[/tex]

With [tex]\alpha/2=0.05[/tex], we can find the value for [tex]z_{\alpha/2}[/tex] using the normal standard distribution table or excel.

The calculated value is [tex]z_{\alpha/2}=1.644854[/tex]

Now from equation (1) we need to solve for n in order to answer the question.

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

Squaring both sides:

[tex](\frac{ME}{z_{\alpha/2}})^2=\frac{\hat p(1-\hat p)}{n}[/tex]

And solving for n we got:

[tex]n=\frac{\hat p(1-\hat p)}{(\frac{ME}{z_{\alpha/2}})^2}[/tex]

Now we can replpace the values

[tex]n=\frac{0.5(1-0.5)}{(\frac{0.04}{1.644854})^2}=422.739[/tex]

And rounded up to the nearest integer we got:

n=423

Answer:

a then b

Step-by-step explanation:

a then b because you could mess up if you didn't draw a decision tree

Correct graph is C

Step-by-step explanation:

Given two inequalities are:

1. [tex]y\leq -3x+1[/tex]

2.[tex]y\leq x+3[/tex]

Step1 :  Remove the inequalities

1. [tex]y =-3x+1[/tex]

2.[tex]y = x+3[/tex]

Step2 :  Finding intersection points of equations

By solving linear equation

[tex]y =-3x+1 =x+3 [/tex]

[tex]-3x+1 =x+3 [/tex]

[tex] x = -0.5[/tex]

Replacing value of x in any equations

we get,

[tex]y = x+3[/tex]

[tex]y =-0.5+3[/tex]

[tex]y = 2.5[/tex]

Therefore, Point of intersection is (-0.5,2.5)

Step3: Test of origin (0,0)

Here, If inequalities holds true for origin then, shades the graph towards the origin.

For equation 1.

[tex]y\leq -3x+1[/tex]

[tex]0\leq -3(0)+1[/tex]

[tex]0\leq +1[/tex]

True, Shade graph towards origin.

For equation 2.

[tex]y\leq x+3[/tex]

[tex]0\leq 0+3[/tex]

[tex]0\leq 3[/tex]

True, Shade graph towards origin.

Thus, Correct graph is C

Answer:

c

Step-by-step explanation:

took test

Answer:

  190

Step-by-step explanation:

C(n, k) = n!/(k!(n -k)!)

C(190, 1) = 190!/(1!(189!)) = 190/1 = 190

The number of combinations of 190 things taken 1 at a time is 190.

Answer:

Step-by-step explanation:

Hello!

You have the following experiment, a random sample of 106 people was made and they were asked if they dreamed in color. 80 persons of the sample reported dreaming in color.

The historical data from the 1940s informs that the population proportion of people that dreams in color is 0.21(usually when historical data is given unless said otherwise, is considered population information)

Your study variable is a discrete variable, you can define as:

X: Amount of people that reported dreaming in colors in a sample of 106.

Binomial criteria:

1. The number of observation of the trial is fixed (In this case n = 106)

2. Each observation in the trial is independent, this means that none of the trials will have an effect on the probability of the next trial (In this case, the fact that one person dreams in color doesn't affect or modify the probability of the next one dreaming in color)

3. The probability of success in the same from one trial to another (Or success is dreaming in color and the probability is 0.21)

So X~Bi(n;ρ)

In order to be able to run a proportion Z-test you have to apply the Central Limit Theorem to approximate the distribution of the sample proportion to normal:

^ρ ≈ N(ρ; (ρ(1-ρ))/n)

With this approximation, you can use the Z-test to run the hypothesis.

Now what the investigators want to know is if the proportion of people that dreams in color has change since the 1940s, so the hypothesis is:

H₀: ρ = 0.21

H₁: ρ ≠ 0.21

α: 0.10

The test is two-tailed,

Left critical value: [tex]Z_{\alpha/2} = Z_{0.95} = -1.64[/tex]

Right critical value: [tex]Z_{1-\alpha /2} = Z_{0.95} = 1.64[/tex]

If the calculated Z-value ≤ -1.64 or ≥ 1.64, the decision is to reject the null hypothesis.

If -1.64 < Z-value < 1.64, then you do not reject the null hypothesis.

The sample proportion is ^ρ= x/n = 80/106 = 0.75

Z=     0.75 - 0.21     = 12.83

    √[(0.75*0.25)/106]

⇒Decision: Reject the null hypothesis.

p-value < 0.00001 is less than 0.10

If you were to conduct a one-tailed upper test (H₀: ρ = 0.21 vs H₁: ρ > 0.21) with the information of this sample, at the same level 10%, the critical value would be [tex]Z_{0.90}[/tex]1.28 against the 12.83 from the Z-value, the decision would be to reject the null hypothesis (meaning that the proportion of people that dreams in colors has increased.)

I hope this helps!

Answer:

[tex]\mu_{\hat{p}}=0.19[/tex]

[tex]\sigma_{\hat{p}}=0.0785[/tex]

Step-by-step explanation:

We know that the mean and the standard error of the sampling distribution of the sample proportions will be :-

[tex]\mu_{\hat{p}}=p[/tex]

[tex]\sigma_{\hat{p}}=\sqrt{\dfrac{p(1-p)}{n}}[/tex]

, where p=population proportion and n= sample size.

Given : The proportion of students at a college who have GPA higher than 3.5 is 19%.

i.e. p= 19%=0.19

The for sample size n= 25

The mean and the standard error of the sampling distribution of the sample proportions will be :-

[tex]\mu_{\hat{p}}=0.19[/tex]

[tex]\sigma_{\hat{p}}=\sqrt{\dfrac{0.19(1-0.19)}{25}}\\\\=\sqrt{0.006156}=0.0784601809837\approx0.0785[/tex]

Hence , the mean and the standard error of the sampling distribution of the sample proportions :

[tex]\mu_{\hat{p}}=0.19[/tex]

[tex]\sigma_{\hat{p}}=0.0785[/tex]

Answer: No, these response does not provide strong evidence that the 34% figure is not accurate for this region.

Step-by-step explanation:

Since we have given that

p = 0.34

x= 228

n = 750

So, [tex]\hat{p}=\dfrac{x}{n}=\dfrac{228}{750}=0.304[/tex]

So, hypothesis would be

[tex]H_0:p=\hat{p}\\\\H_a:p\neq \hat{p}[/tex]

So, test statistic value would be

[tex]z=\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}\\\\z=\dfrac{0.304-0.38}{\sqrt{\dfrac{0.38\times 0.62}{750}}}\\\\z=\dfrac{-0.076}{0.0177}\\\\z=-4.293[/tex]

At 95% confidence , z = 1.96

So, 1.96>-4.293.

So, we accept the null hypothesis.

No, these response does not provide strong evidence that the 34% figure is not accurate for this region.

The observed data doesn't provide enough evidence to suggest that the 34% figure reported by the CDC is inaccurate for this region.

let's break down the hypothesis test step by step with more detail.

1. Setting up the Hypotheses:

- Null Hypothesis (H0):The true proportion of obese adults in the region is 34%.

- Alternative Hypothesis (H1):The true proportion of obese adults in the region is not 34%.

2. Calculating the Expected Number of Obese Individuals:

To calculate the expected number of obese individuals in the sample under the assumption that the true proportion is 34%, we use the formula:

[tex]\[ \text{Expected number of obese individuals} = \text{Proportion} \times \text{Sample size} \][/tex]

So,

[tex]\[ \text{Expected number of obese individuals} = 0.34 \times 750 = 255 \][/tex]

3. Checking Conditions:

- The sample is randomly selected.

- The sample size is large enough for the Central Limit Theorem to apply (n = 750).

- Since the population size isn't provided, we'll assume it's much larger than the sample size (which is often the case with populations of adults in countries).

4. Calculating the Z-Score:

The formula for the z-score is:

[tex]\[ z = \frac{{\text{Observed proportion} - \text{Expected proportion}}}{{\sqrt{\frac{{\text{Expected proportion} \times (1 - \text{Expected proportion})}}{{\text{Sample size}}}}}} \][/tex]

So,

[tex]\[ z = \frac{{\frac{228}{750} - \frac{255}{750}}}{{\sqrt{\frac{255}{750} \times \frac{495}{750}}}} \]\[ z \approx \frac{{0.304 - 0.34}}{{\sqrt{\frac{191.25}{750}}}} \]\[ z \approx \frac{{-0.036}}{{\sqrt{0.255}}} \]\[ z \approx \frac{{-0.036}}{{0.505}} \]\[ z \approx -0.071 \][/tex]

5. Finding Critical Z-Value:

For a two-tailed test with a significance level of 0.05, the critical z-values are approximately ±1.96.

6. Making a Decision:

Since the calculated z-score (-0.071) falls within the range (-1.96, 1.96), we fail to reject the null hypothesis. This means there is not enough evidence to conclude that the true proportion of obese adults in the region is different from 34%.

In conclusion, based on the provided sample data, we cannot confidently claim that the 34% figure reported by the CDC is inaccurate for this region.

Answer:

1.25 liters of oil

Step-by-step explanation:

Volume in Beaker A = 1 L

Volume of Oil in Beaker A = 1*0.3 = 0.3 L

Volume of Vinegar in Beaker A = 1*0.7 = 0.7 L

Volume in Beaker B = 2 L

Volume of Oil in Beaker B = 2*0.4 = 0.8 L

Volume of Vinegar in Beaker B = 1*0.6 = 1.2 L

If half of the contents of B are poured into A and assuming a homogeneous mixture, the new volumes of oil (Voa) and vinegar (Vva) in beaker A are:

[tex]V_{oa} = 0.3+\frac{0.8}{2} \\V_{oa} = 0.7 \\V_{va} = 0.7+\frac{1.2}{2} \\V_{va} = 1.3[/tex]

The amount of oil needed to be added to beaker A in order to produce a mixture which is 60 percent oil (Vomix) is given by:

[tex]0.6*V_{total} = V_{oa} +V_{omix}\\0.6*(V_{va}+V_{oa} +V_{omix}) = V_{oa} +V_{omix}\\0.6*(1.3+0.7+V_{omix})=0.7+V_{omix}\\V_{omix}=\frac{0.5}{0.4} \\V_{omix}=1.25 \ L[/tex]

1.25 liters of oil are needed.

There is 1.25 litres of oil that should now be added to A to produce a mixture that is 60 per cent oil.

Given

Beaker A contains 1 litre which is 30 per cent oil and the rest is vinegar, thoroughly mixed up.

Beaker B contains 2 litres which are 40 per cent oil and the rest vinegar, completely mixed up.

Half of the contents of B are poured into A, then completely mixed up.

Contents in beaker A implies;

15% of oil in 1 litre = 0.15 litre of oil

So that, there are 0.15 litres of oil and 0.85 litres of vinegar in beaker A.

Contents in beaker B implies:

55% of oil in 2 litres = 1.1 litres of oil

So that, thee are 1.1 litres of oil and 0.9 litres of vinegar in beaker B.

Half of the contents of B poured into A implies that beaker A now contains:

0.15 litres + 0.55 litres = 0.7 litres of oil

0.85 litres + 0.45 litres = 1.3 litres of vinegar

Then,

The percentage of oil in A is;

[tex]=\dfrac{0.7}{2} \times 100\\\\= 35[/tex]

To increase the percentage of oil to 60%, then:

0.7 litres + 1.25 litres = 1.95 litres of oil

And The new total litres of the content in beaker A = 3.25 litres

[tex]=\dfrac{1.95}{3.25}\times 100\\\\=60 \rm \ percent[/tex]

Hence, 1.25 litres of oil should now be added to A to produce a mixture that is 60 per cent oil.

To know more about percentages click the link given below.

https://brainly.com/question/12721886?

Answer: 10

Step-by-step explanation:

We know that the formula to find the sample size is given by :-

[tex]n= (\dfrac{z_{\alpha/2\cdot \sigma}}{E})^2[/tex]

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

[tex]z_{\alpha/2}[/tex] = Two -tailed z-value for [tex]{\alpha[/tex] (significance level)

E= margin of error.

Given : Confidence level : C =99%=0.99

i.e. [tex]1-\alpha=0.99[/tex]

⇒Significance level :[tex]\alpha=1-0.99=0.01[/tex]

By using z-value table ,Two -tailed z-value for [tex]\alpha=0.01 [/tex]:

[tex]z_{\alpha/2}=2.576[/tex]

E= 2 minutes

[tex]\sigma=\text{2.4 minutes}[/tex]

The required sample size will be :-

[tex]n= (\dfrac{2.576\cdot 2.4}{2})^2\\\\= (3.0912)^2\\\\=9.55551744\approx10[/tex]

Hence, the required sample size = 10

To estimate the mean time college students spend watching online videos each day within 2 minutes of the population mean at a 99% confidence level, at least 10 students should be sampled assuming a population standard deviation of 2.4 minutes.

To determine the required sample size, we use the formula for the sample size of a mean:

n = (z*σ/E)^2

Where n is the sample size, z is the z-score corresponding to the desired confidence level, σ is the population standard deviation, and E is the maximum error allowed (margin of error).

For a 99% confidence interval, the z-score (from the standard normal distribution) is approximately 2.576. The population standard deviation σ is given as 2.4 minutes, and the margin of error E is 2 minutes. Using these values:

n = (2.576*2.4/2)^2

n ≈ (6.1824/2)^2

n ≈ (3.0912)^2

n ≈ 9.554

Since the sample size must be a whole number, we would round up to ensure the sample size is large enough to achieve the desired margin of error, which means at least 10 students should be surveyed to construct the 99% confidence interval for the mean time spent watching online videos.

It is important to note that the larger the sample size, the more accurate the estimate of the population mean will be.

Answer:

Initial velocity    192.666 ft/sec

a) 1.41 sec

Step-by-step explanation:

Equations for

V(f)  = V₀ ± at

V(f)²  =( V₀)² ± 2 a*d

d =   V₀*t  ±  a*t²/2

We are going to use a(t)  =  g  = 32 ft/sec²

a) V₀   =  ??         a = -g      

Then   as  d = 580 ft       and  V(f)  =  0

we have

V(f)²  =  ( V₀)²  - 2*32*580      ⇒  ( V₀)²   = 64*580  (ft²/sec²)

( V₀)²   =  37120          V₀  = 192,666 ft/sec      

a)  t  = ??   in this case V₀ = 0

d  =    V₀*t + gt²/2       ⇒    32  =   ( 32 t²)  /2

t²  = 2  

t =   1.41 sec

Answer:

  [tex]\dfrac{1}{x^2}+\dfrac{y^2}{x^4}[/tex]

Step-by-step explanation:

Separate the fraction(s) at the plus sign and simplify.

[tex]\dfrac{x^2+y^2}{x^4}=\dfrac{x^2}{x^4}+\dfrac{y^2}{x^4}\\\\=\dfrac{1}{x^2}+\dfrac{y^2}{x^4}[/tex]

Answer:

The answer is  b

Step-by-step explanation:

cause we have the equation y- y1=m(x-x1)

so y-3=4/11(x-11)

Final answer:

The volume of the pancreas is estimated using the Midpoint Rule by adding the cross-sectional areas and multiplying by the distance between sections, resulting in an estimated volume of 100.5 cubic centimeters.

Explanation:

The question asks us to use the Midpoint Rule to estimate the volume of the human pancreas given its length (12 cm) and cross-sectional areas at 1 cm intervals. To do this, we add up the areas of the cross-sections (excluding the first and last since they're 0) and multiply by the distance between the sections (1 cm). This approach approximates the volume using cylindrical segments where each segment's volume is its cross-sectional area times the height (1 cm).

The given cross-sectional areas are 7.9, 15.3, 18.1, 10.2, 10.7, 9.5, 8.5, 7.8, 5.6, 4.2, and 2.7 square cm. Adding these gives a total of 100.5 square cm. Since the distance between each section is 1 cm, multiplying 100.5 by 1 cm gives an estimated pancreatic volume of 100.5 cubic centimeters.

This method, while not perfectly accurate due to it being an approximation, provides a useful estimate of the volume based on the available data. It's a practical application of the Midpoint Rule in estimating volumes from cross-sectional data.

Answer:

Step-by-step explanation:

The equation for the number of can tabs y that she has collected is

y = 4x + 35

where x is the number of weeks. To plot the graph, values for different number of weeks are inputted to get the corresponding y values.

Comparing the equation with the slope intercept form,

y = mx + c,

m represents slope.

So slope of the equation is 4. It represents the rate at which the total number of can tabs that she has collected is increasing with respect to the increase in the number of weeks. So the total number increases by 4 each week

The y intercept is the point at x = 0

It is equal to 35. It represents the number of can tabs that she had initially collected before she started collecting 4 can tabs weekly. At this point, her weekly collection is 0

a. Using Hamilton's method, 4 counselors should be assigned to Lowell, and 9 counselors should be assigned to Fairview.

b. The same divisor, 2 counselors should be hired for the new school.

Hamilton's method is used to allocate resources, such as counselors, among several entities based on a certain divisor. The divisor is typically calculated by dividing the total number of students by the total number of counselors available.

Let's start by determining the divisor and assigning counselors to each school based on the given information:

1. Calculate the divisor:

[tex]\[ \text{Divisor} = \frac{\text{Total Students}}{\text{Total Counselors}} \]\\\text{Divisor} = \frac{3584 + 6816}{13} = \frac{10400}{13} \approx 800 \][/tex]

2. Assign counselors to each school:

[tex]\[ \text{Counselors Assigned to Lowell} = \frac{\text{Students Enrolled in Lowell}}{\text{Divisor}} \][/tex]

[tex]\[ \text{Counselors Assigned to Fairview} = \frac{\text{Students Enrolled in Fairview}}{\text{Divisor}} \][/tex]

[tex]\[ \text{Counselors Assigned to Lowell} = \frac{3584}{800} = 4.48 \approx 4 \][/tex]

[tex]\[ \text{Counselors Assigned to Fairview} = \frac{6816}{800} = 8.52 \approx 9 \][/tex]

So, using Hamilton's method, 4 counselors should be assigned to Lowell, and 9 counselors should be assigned to Fairview.

3. New School:

  If a new school with 1824 students is opened, we can use the same divisor to determine how many additional counselors should be hired for the new school:

[tex]\[ \text{Counselors for New School} = \frac{\text{Students in New School}}{\text{Divisor}} \][/tex]

[tex]\[ \text{Counselors for New School} = \frac{1824}{800} = 2.28 \approx 2 \][/tex]

So, using the same divisor, 2 counselors should be hired for the new school.

Answer:

a) The probability that a pregnancy will last 308 days or longer is 0.0038

b) Babies who are born on or before 256 days are considered prematures.

Step-by-step explanation:

Let X be the random variable that represents the length of a pregnancy. Then, X is normally distributed with a mean of 268 days and a standard deviation of 15 days.  

a) The z-score related to 308 days is z = (308-268)/15 = 2.6667, so, the probability of a pregnancy lasting 308 days or longer is P(Z > 2.6667) = 0.0038

b) We are looking for a value q such that P(X < q) = 0.22, i.e., P((X-268)/15 < (q-268)/15) = 0.22, here, Z =  (X-268)/15 is a standard normal random variable and z = (q-268)/15 is the 22nd quantile of the standard normal distribution, i.e., z = -0.7722 =  (q-268)/15 and (-0.7722)(15) + 268 = q, i.e., q = 256.417, so, babies who are born on or before 256 days are considered premature.

Answer:

a) P(X>308) = 0.00383

b) 256.42 days

Step-by-step explanation:

Population mean (μ) = 268 days

Standard deviation (σ) = 15 days

a) P(X>308)?

The z-score for any length of pregnancy 'X' is given by:

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

The z-score for X=308 is:

[tex]z=\frac{308-268}{15}\\z=2.6667[/tex]

A z-score of 2.6667 is equivalent to the 99.617-th percentile of a normal distribution. Thus, the probability of a pregnancy lasting 208 days or longer is:

[tex]P(X>308)=1-0.99617\\P(X>308) = 0.00383[/tex]

b) X at which P(X) < 0.22?

At the 22-nd percentile, a normal distribution has an equivalent z-score of -0.772. Therefore, the length of pregnancy, X, that separates premature babies from those who are not premature is:

[tex]-0.772=\frac{X-268}{15}\\X= 256.42[/tex]

Answer:

1044 sq. inches

Step-by-step explanation:

The frame adds 2 inches on each side of the window, so the width becomes 29 in. and the length becomes 36 in.

Area = width × length

So the total area of the window and the frame is:

A = 29 × 36 = 1044 in²

Answer:

D. Null Hypothesis:μ=25 Alt Hypothesis:μ < 25

Step-by-step explanation:

Previous concepts and notation

A hypothesis is defined as "a speculation or theory based on insufficient evidence that lends itself to further testing and experimentation. With further testing, a hypothesis can usually be proven true or false".  

The null hypothesis is defined as "a hypothesis that says there is no statistical significance between the two variables in the hypothesis. It is the hypothesis that the researcher is trying to disprove".

The alternative hypothesis is "just the inverse, or opposite, of the null hypothesis. It is the hypothesis that researcher is trying to prove".

[tex]\bar X= 23.9[/tex] represent the sample mean

[tex]s=3[/tex] represent the sample standard deviation

n=25 represent the sample selected

Solution

The claim that we want to test is that: "You believe that the mean BMI for college students is less than 25". So this statement needs to be on the Alternative hypothesis ([tex]\mu <25[/tex]) and the Null hypothesis would be the complement ([tex]\mu =25[/tex]) or ([tex]\mu \geq 25[/tex]).

So the best option on this case is:

D. Null Hypothesis:μ=25 Alt Hypothesis:μ < 25

Answer:

a) By the central limit theorem we know that if the random variable X="quantitative scores for young women", and we know that this distribution is normal.

[tex]X \sim N(\mu, \sigma=60)[/tex]

And we are interested on the distribution for the sample mean [tex]\bar X[/tex], we know that distribution for the sample mean is given by:

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

And with that we have the assumptions required to apply the normal distribution to create the confidence interval since the distribution for [tex]\bar X[/tex] is normal.

b) [tex]270.283\leq \mu \leq 279.717[/tex]

c) We are confident (99%) that the true mean for the quantitative scores for young women is between (270.283;3279.717)

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

[tex]\bar X=275[/tex] represent the sample mean for the sample  

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

[tex]\sigma=60[/tex] represent the population standard deviation

n=1077 represent the sample size  

Part a

By the central limit theorem we know that if the random variable X="quantitative scores for young women", and we know that this distribution is normal.

[tex]X \sim N(\mu, \sigma=60)[/tex]

And we are interested on the distribution for the sample mean [tex]\bar X[/tex], we know that distribution for the sample mean is given by:

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

And with that we have the assumptions required to apply the normal distribution to create the confidence interval since the distribution for [tex]\bar X[/tex] is normal.

Part b

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

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

Since the Confidence is 0.99 or 99%, the value of [tex]\alpha=0.01[/tex] and [tex]\alpha/2 =0.005[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.005,0,1)".And we see that [tex]z_{\alpha/2}=2.58[/tex]

Now we have everything in order to replace into formula (1):

[tex]275-2.58\frac{60}{\sqrt{1077}}=270.283[/tex]    

[tex]275+2.58\frac{60}{\sqrt{1077}}=279.717[/tex]

So on this case the 99% confidence interval would be given by (270.283;3279.717)    

[tex]270.283\leq \mu \leq 279.717[/tex]

Part c

We are confident (99%) that the true mean for the quantitative scores for young women is between (270.283;3279.717)

Final answer:

A 99% confidence interval for the mean quantitative scores for young women, based on the NAEP data, is found to be between 270.27 and 279.73. We can state with 99% confidence that the true mean quantitative score for young women falls within this range.

Explanation:

We are asked to find a 99% confidence interval for the mean quantitative scores for young women, based on the National Assessment of Educational Progress (NAEP) sample data.

a) Check that the normality assumptions are met.

The normality assumption is met because individual NAEP scores are given to have a Normal distribution. Additionally, with a large sample size (more than 30), the Central Limit Theorem ensures the sampling distribution of the mean is approximately normal, regardless of the distribution of the population from which the sample is drawn.

b) What is the 99% confidence interval for the mean quantitative scores for young women?

Using the given data, the sample mean, \\bar{x}\, is 275 and the population standard deviation, \(\sigma\), is 60. Since the population standard deviation is given and the sample size is large (1077), we will use the z-distribution to find the 99% confidence interval.

The z-score corresponding to a 99% confidence level is approximately 2.576. Therefore, the margin of error (ME) can be calculated as:

ME = z * [tex](\(\sigma/\sqrt{n}\))[/tex]

ME = 2.576 * [tex](60/\sqrt{1077})[/tex] \approx 4.73

The 99% confidence interval is:

Lower limit =[tex]\(\bar{x}\)[/tex] - ME = 275 - 4.73 = 270.27

Upper limit =[tex]\(\bar{x}\)[/tex] + ME = 275 + 4.73 = 279.73

The 99% confidence interval for the mean quantitative scores for young women is 270.27 ≤ μ ≤ 279.73.

c) Interpret the confidence interval obtained in previous question

The interpretation of this confidence interval is that we are 99% confident that the true mean quantitative score for young women falls between 270.27 and 279.73.

The probability of a pug approaching first in a park of 25 dogs is 5/25 or 1/5. The average grade of a larger class is typically a more reliable representation of the group's performance than that of a smaller one.

For the first question, the probability of a pug being the first dog to approach you in a park with 25 dogs, 5 of them being pugs, is indeed 5/25, or 1/5. This fraction represents the ratio of the desired outcome (a pug approaching you first) to the total number of possible outcomes (any of the 25 dogs in the park could potentially be the first to approach).

For the second part of the question - determining the more reliable average grade between two classes - the class with 50 students is the better choice. This conclusion is based on statistical reasoning. A larger sample size (in this case, the larger class) generally provides a more reliable average than a smaller sample size.

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Answer:

a. purple to orange  =[tex]\frac{5}{6}[/tex]

b. purple to all beads=[tex]\frac{5){11}[/tex]

c. all beads to orange=[tex]\frac{11}{6}[/tex]

Step-by-step explanation:

Given:

Colour of beads in jar = orange or purple

The ratio of orange to purple beads = 6 : 5

Solution:

Let number of orange beads be= 6x

Number of purple  beads=5x

Total no of beads= 6x+5x=11x

a. Ratio of  purple to orange.

Ratio of  purple to orange=[tex]\frac{\text{number of purple boads}}{\text{number of orange beads}}[/tex]

Ratio of  purple to orange=[tex]\frac{5x}{6x}[/tex]

Ratio of  purple to orange=[tex]\frac{5}{6}[/tex]

b. Ratio of purple to all beads

Ratio of  purple to all beads=[tex]\frac{\text{number of purple beads}}{\text{Total number of  beads}}[/tex]

Ratio of purple to all beads=[tex]\frac{5x}{11x}[/tex]

Ratio of purple to all beads=[tex]\frac{5}{11}[/tex]

c. Ratio of all beads to orange

Ratio of  purple to all beads=[tex]\frac{\text{Total number of  beads}}{\text{number of orange beads}}[/tex]

Ratio of purple to all beads=[tex]\frac{11x}{6x}[/tex]

Ratio of purple to all beads=[tex]\frac{11}{6}[/tex]

Answer:

0.082

Step-by-step explanation:

Number of attempts = n = 100

Since there are only two outcomes and in-dependent of each other, the probability of missing a shot  = 1 - Probability of making each shot

p = 1 - 0.95 = 0.05

Possion Ratio (λ) = np where n is the number of events and p is the probability of the shot missing

λ = 100 x 0.05 = 5

Define X such that X = Number of misses and X ≅ Poisson (λ = 5)

P [X ≤ 2] = P [X = 0] + P [X = 1] + P [X = 2]

P [X ≤ 2] = e⁻⁵ + e⁻⁵ x 5 + e⁻⁵ x 5²/2!

P [X ≤ 2] = e⁻⁵ [1 + 5 + 5²/2!]

P [X ≤ 2] = e⁻⁵ x 12.25 = 0.082

The required probability that there are at most 2 misses in the first 100 attempts is 0.082

Answer:

57,600,000

Step-by-step explanation:

Answer:

a) New mean 78.07692308 thousands of dollars

b) The median does not vary

c) New standard deviation 150.1793799 thousands of dollars

Step-by-step explanation:

We are working in units of $1,000

a)What is the value of the mean after the change?

Let  

[tex]s_1,s-2,...,s_38[/tex]  

be the salaries of the employees that earn less than 100 units.

The mean of the 39 salaries is 55 units so

[tex]\displaystyle\frac{s_1+s_2+...+s_{38}+100}{39}=55[/tex]

and

[tex]s_1+s_2+...+s_{38}+100=55*39=2145[/tex]

By accident, the 100 on the left is changed to 1,000

[tex]s_1+s_2+...+s_{38}+100+900=55*39=2145+900\Rightarrow \\\\\Rightarrow s_1+s_2+...+s_{38}+1000=3045[/tex]

Dividing by 39 both sides, we get the new mean

[tex]\displaystyle\frac{s_1+s_2+...+s_{38}+1000}{39}=\displaystyle\frac{3045}{39}=78.07692308[/tex]

b)What is the value of the median after the change?

Since the number of data does not change and only the right end of the range of salaries is changed, the median remains the same; 55

c)What is the value of the standard deviation after the change?

The variance is 400, so

[tex]\displaystyle\frac{(s_1-70)^2+(s_2-70)^2+...+(s_{38}-70)^2+(100-70)^2}{39}=400\Rightarrow\\\\\Rightarrow (s_1-70)^2+(s_2-70)^2+...+(s_{38}-70)^2+(100-70)^2=400*39=15600[/tex]

Adding 864,000 to both sides we get

[tex](s_1-70)^2+(s_2-70)^2+...+(s_{38}-70)^2+(100-70)^2+864000=15600+864000\Rightarrow\\\\\Rightarrow (s_1-70)^2+(s_2-70)^2+...+(s_{38}-70)^2+(1000-70)^2=879600[/tex]

Dividing by 39 and taking the square root we get the new standard deviation

[tex]\displaystyle\frac{(s_1-70)^2+(s_2-70)^2+...+(s_{38}-70)^2+(1000-70)^2}{39}=\displaystyle\frac{879600}{39}=22553.84615\Rightarrow\\\\\Rightarrow \sqrt{\displaystyle\frac{(s_1-70)^2+(s_2-70)^2+...+(s_{38}-70)^2+(1000-70)^2}{39}}=150.1793799[/tex]

Final answer:

a) The new mean after the change is approximately $66,923,000. b) The new median after the change is approximately $60,961,000. c) The new standard deviation after the change is approximately $234,358,000.

Explanation:

a) To find the new value of the mean, we need to subtract the original largest number ($100,000) and add the new largest number ($1,000,000) to the sum of the salaries. The sum of the salaries can be calculated by multiplying the mean ($70,000) by the number of employees (39). So, the new sum of the salaries is $(70,000 × 39) - $100,000 + $1,000,000 = $2,610,000. Since there are still 39 employees, the new mean is $2,610,000 divided by 39, which is approximately $66,923. The value of the mean after the change is $66,923,000 (in units of $1000).

b) The median is the middle value in a list of numbers when they are ordered from smallest to largest. After the change, the values in the list would be: $55,000, $55,000, $55,000, ..., $55,000, $66,923, $66,923, ..., $1,000,000. Since there are 39 employees, the middle two values would be $55,000 and $66,923. To find the median, we take the average of these two values: ($55,000 + $66,923) / 2 = $60,961. The value of the median after the change is $60,961,000 (in units of $1000).

c) To find the new standard deviation, we need to recalculate it using the new values. First, we need to find the squared differences between each salary and the new mean. The sum of these squared differences is $(39 × ($66,923 - $66,923)^2) + $1,000,000. Then, we divide this sum by 39 and take the square root to find the new standard deviation. The value of the standard deviation after the change is approximately $234,358,000 (in units of $1000).

Answer:

(a) Hypotheses relevant to the research question are  

[tex]H_{0}: \mu_{1}-\mu_{2} = 0[/tex] vs [tex]H_{1}: \mu_{1}-\mu_{2} < 0[/tex] (lower-tail alternative) and

[tex]H_{0}: \mu_{1}-\mu_{2} = 0[/tex] vs [tex]H_{1}: \mu_{1}-\mu_{2} > 0[/tex] (upper-tail alternative).  

(b) There is no evidence that one mathematic learning technique is better than the other.

Step-by-step explanation:

Let's suppose that the data related to Technique A comes from population 1 and that the data related to Technique B comes from population 2. We have large sample sizes [tex]n_{1} = 90[/tex]  and [tex]n_{2} = 110[/tex]. The unbiased point estimate for [tex]\mu_{1}-\mu_{2}[/tex] is [tex]\bar{x}_{1} - \bar{x}_{2}[/tex], i.e., 26.1 - 27.9 = -1.8. The standard error is given by [tex]\sqrt{\frac{\sigma^{2}_{1}}{n_{1}}+\frac{\sigma^{2}_{2}}{n_{2}}}\approx[/tex] [tex]\sqrt{\frac{(7.2)^{2}}{90}+\frac{(12.4)^{2}}{110}}[/tex] = 1.4049.

(a) Hypotheses relevant to the research question are  

[tex]H_{0}: \mu_{1}-\mu_{2} = 0[/tex] vs [tex]H_{1}: \mu_{1}-\mu_{2} < 0[/tex] (lower-tail alternative) and

[tex]H_{0}: \mu_{1}-\mu_{2} = 0[/tex] vs [tex]H_{1}: \mu_{1}-\mu_{2} > 0[/tex] (upper-tail alternative).  

(b) 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{-1.8}{1.4049} = -1.2812[/tex].  

The p-value for the lower-tail alternative is P(Z < -1.2812) =  0.1000617 and the p-value for the upper-tail alternative is P(Z > -1.2812) = 0.8999. The p-value is greater than 0.10 in both cases, therefore, we fail to reject the null hypotheses in both cases.

The null hypothesis states that the mean completion time using Technique A equals that using Technique B. The alternative hypothesis states that they are not equal. A t-test can be used to test these hypotheses. The test statistic and the P-value are calculated and compared with the given significance level (0.10) to make a conclusion.

Given the data provided for Techniques A and B, let’s formulate the hypothesis and carry out the test to determine the effectiveness of the techniques.

(a) State the Hypotheses

Here, the null hypothesis (H₀) is that the mean completion time using Technique A equals that using Technique B. The alternative hypothesis (H₁) is that the mean time using Technique A is not equal to that using Technique B.

(b) Hypothesis Testing

For the hypothesis test, we have to calculate the t-statistic and the P-value. Given that the data is normally distributed, we can use the two-sample t-test. The standard formula used to calculate the t-statistic is: (sample mean₁ - sample mean₂) / √[(sample variance₁/n₁) + (sample variance₂/n₂)]. After calculating these values, the resultant t-statistic can be compared to critical t values for the given level of significance (α=10%).

Next, compute the P-value using the t-statistic, which indicates the probability of observing such a difference in means under the null hypothesis. We reject the null hypothesis if the P-value is less than our significance level (α).

Without specific test statistics and P-value, we can’t make a definitive conclusion. In general, if we have a statistically significant result (P<0.10), we would reject the null hypothesis and conclude that there is a significant difference in the performance of Techniques A and B.

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Answer: The value of test statistic is 0.696.

Step-by-step explanation:

Since we have given that

[tex]n_1=1399\\\\x_1=411\\\\p_1=\dfrac{x_1}{n_1}=\dfrac{411}{1399}=0.294[/tex]

Similarly,

[tex]n_2=759, x_2=211\\\\p_2=\dfrac{x_2}{n_2}=\dfrac{211}{759}=0.278[/tex]

At 0.01 level of significance.

Hypothesis are :

[tex]H_0:P_1=P_2=0.5\\\\H_a:P_1\neq P_2[/tex]

So, the test statistic value would be

[tex]z=\dfrac{(p_1-p_2)-(P_1-P_2)}{\sqrt{\dfrac{P_1Q_1}{n_1}+\dfrac{P_2Q_2}{n_2}}}\\\\z=\dfrac{0.294-0.278}{\sqrt{0.5\times 0.5(\dfrac{1}{1399}+\dfrac{1}{759})}}\\\\z=\dfrac{0.016}{0.023}=0.696[/tex]

Hence, the value of test statistic is 0.696.

Answer:

138

Step-by-step explanation:

Population standard deviation = [tex]\sigma = 120[/tex]

.95 probability of estimating the population mean monthly income within a margin of $20

So, Significance level = 1-0.95 = 0.05

α =0.05

Margin error = 20

[tex]ME =Z \times \frac{\sigma}{\sqrt{n}}[/tex]

Z at 0.05 = 1.96

[tex]20 =1.96 \times \frac{120}{\sqrt{n}}[/tex]

[tex]\sqrt{n} =1.96 \times \frac{120}{20}[/tex]

[tex]n =(1.96 \times \frac{120}{20})^2[/tex]

[tex]n =138.2976[/tex]

So, n = 138

Hence sample size should be 138 selected to obtain a .95 probability of estimating the population mean monthly income within a margin of $20