In fall 2014, 36% of applicants with a Math SAT of 700 or more were admitted by a certain university, while 18% with a Math SAT of less than 700 were admitted. Further, 38% of all applicants had a Math SAT score of 700 or more. What percentage of admitted applicants had a Math SAT of 700 or more? (Round your answer to the nearest percentage point.)

Answer:

A. 2.5 standard deviations below the average salary.

Step-by-step explanation:

Hello!

Usually, when you calculate a probability of an X₀ value from a normally distributed variable you need to standardize it to reach the proper value. In this case, you already have the Z-value and need to reverse the standardization to obtain the X₀ value.

Z= X₀ - μ

δ

δ*Z = X₀ - μ

X₀= (δ*Z) + μ

replace Z= -2.50

X₀= (δ*-2.50) + μ

X₀= μ - (δ*2.50)

You can replace the Z value any time. I've just choose to do it at the end because it's more confortable for me.

I hope it helps!

Answer: A bag can weigh 13.77 ounces and not need to be repacked.

Step-by-step explanation:

Since we have given that

Mean = 12.08 ounces

Standard deviation = 1.03 ounces

Bags in the upper 5% are too heavy and must be repackaged.

Using the standard normal table.

z = 1.645

So, [tex]X=\mu+z\sigma\\\\X=12.08+1.645\times 1.03\\\\X=13.77\ ounces[/tex]

Hence, A bag can weigh 13.77 ounces and not need to be repacked.

Answer:

Null hypothesis: [tex]\mu_1 \geq \mu_2[/tex]

Alternative hypothesis: [tex]\mu_1 <\mu_2[/tex]

[tex]t=-1.329[/tex]

[tex]p_v =P(t_{30}<-1.329) =0.0969[/tex]

With the p value obtained and using the significance level given [tex]\alpha=0.1[/tex] we have [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 10% of significance the mena of the group 1 is significantly lower than the mean for the group 2.  

Step-by-step explanation:

When we have two independnet samples from two normal distributions with equal variances we are assuming that

[tex]\sigma^2_1 =\sigma^2_2 =\sigma^2[/tex]

And the statistic is given by this formula:

[tex]t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{S_p\sqrt{\frac{1}{n_1}}+\frac{1}{n_2}}[/tex]

Where t follows a t distribution with [tex]n_1+n_2 -2[/tex] degrees of freedom and the pooled variance [tex]S^2_p[/tex] is given by this formula:

[tex]\S^2_p =\frac{(n_1-1)S^2_1 +(n_2 -1)S^2_2}{n_1 +n_2 -2}[/tex]

This last one is an unbiased estimator of the common variance [tex]\simga^2[/tex]

The system of hypothesis on this case are:

Null hypothesis: [tex]\mu_1 \geq \mu_2[/tex]

Alternative hypothesis: [tex]\mu_1 <\mu_2[/tex]

Or equivalently:

Null hypothesis: [tex]\mu_1 - \mu_2 \geq 0[/tex]

Alternative hypothesis: [tex]\mu_1 -\mu_2<0[/tex]

Our notation on this case :

[tex]n_1 =14[/tex] represent the sample size for group 1

[tex]n_2 =18[/tex] represent the sample size for group 2

[tex]\bar X_1 =565[/tex] represent the sample mean for the group 1

[tex]\bar X_2 =578[/tex] represent the sample mean for the group 2

[tex]s_1=28.9[/tex] represent the sample standard deviation for group 1

[tex]s_2=26.3[/tex] represent the sample standard deviation for group 2

First we can begin finding the pooled variance:

[tex]\S^2_p =\frac{(14-1)(28.9)^2 +(18 -1)(26.3)^2}{14 +18 -2}=753.882[/tex]

And the deviation would be just the square root of the variance:

[tex]S_p=27.457[/tex]

And now we can calculate the statistic:

[tex]t=\frac{(565 -578)-(0)}{27.457\sqrt{\frac{1}{14}}+\frac{1}{18}}=-1.329[/tex]

Now we can calculate the degrees of freedom given by:

[tex]df=14+18-2=30[/tex]

And now we can calculate the p value using the altenative hypothesis:

[tex]p_v =P(t_{30}<-1.329) =0.0969[/tex]

So with the p value obtained and using the significance level given [tex]\alpha=0.1[/tex] we have [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 10% of significance the mena of the group 1 is significantly lower than the mean for the group 2.  

Answer:

Please see attachment

Step-by-step explanation:

The probability of the sample mean being less than 354.8 ml can be calculated using the z-score and the standard normal distribution. For the second part, if the standard deviation is indeed 4, it's necessary to calculate the probability for a single bottle first, then apply the binomial distribution to find the probability for 10 out of 32 bottles.

The calculation of the probability that the sample mean is less than 354.8 ml when the population mean is 355 ml and the standard deviation is 2 ml for a sample size of 32 uses the standard normal distribution. Applying the Central Limit Theorem, we can find the z-score and use it to calculate the probability.

For part b, assuming the normal distribution N(355, 4), to find the probability that 10 out of 32 bottles contain less than 354.8 ml, we would use the binomial distribution with the probability calculated in part a. However, the second part of question b appears to have a typo with the standard deviation. If the distribution is N(355, 4), we would need to first find the probability of a single bottle having less than 354.8 ml, then use the binomial formula to find the probability of 10 bottles having less than that amount.

Final answer:

To find the equation of the regression line, calculate the slope and y-intercept using the given formulas with the given data points.

Explanation:

To find the equation of the regression line, we need to calculate the slope and the y-intercept. The slope, b, can be found using the formula:

b = (nΣ(xy) - ΣxΣy) / (nΣ(x^2) - (Σx)^2)

where n is the number of data points, Σ represents summation, x and y are the given data points. The y-intercept, a, can be calculated using the formula:

a = (Σy - bΣx) / n

Substituting the values from the given data into these formulas will give you the equation of the regression line.

The researcher would need a sample size of 271 individuals to be 90% confident that the sample proportion will not differ from the true proportion of adults

How to find the sample size ?

To determine the sample size needed to estimate the proportion of the adult population of the United States with high blood pressure with a 90% confidence level and a margin of error (precision) of 5%, you can use the formula for sample size for estimating proportions. The formula is:

[tex]\[n = \frac{Z^2 \cdot p(1-p)}{E^2}\][/tex]

E is the margin of error as a proportion (in this case, 5% or 0.05).

Plug in the values:

[tex]\(n = \frac{(1.645)^2 \cdot 0.5(1-0.5)}{(0.05)^2}\)[/tex]

[tex]\(n = \frac{2.706025 \cdot 0.25}{0.0025}\)[/tex]

[tex]\(n = \frac{0.67650625}{0.0025}\)[/tex]

n = 270.6025

= 271 people

Answer:

(1) 0.4207

(2) 0.7799

Step-by-step explanation:

Given,

Mean value,

[tex]\mu = 15.0[/tex]

Standard deviation,

[tex]\sigma = 1.25[/tex]

(1) P(X ≥ 17.5) = 1 - P( X ≤ 17.5)

[tex]= 1- P(\frac{x-\mu}{\sigma} \leq \frac{17.5-\mu}{\sigma})[/tex]

[tex]=1-P(z\leq \frac{17.5 - 15}{1.25})[/tex]

[tex]=1-P(z\leq \frac{2.5}{1.25})[/tex]

[tex]=1-P(z\leq 2)[/tex]

[tex]=1- 0.5793[/tex]   ( By using z-score table )

= 0.4207

(2) P(14 ≤ X ≤ 18) = P(X ≤ 18) - P(X ≤ 14)

[tex]=P(z\leq \frac{18 - 15}{1.25}) - P(z\leq \frac{14 - 15}{1.25})[/tex]

[tex]=P(z\leq \frac{3}{1.25}) - P(z\leq -\frac{1}{1.25})[/tex]

[tex]=P(z\leq 2.4) - P(z\leq -0.8)[/tex]

= 0.9918 - 0.2119

= 0.7799

The probabiliity that the force is greater than 17.5 kips is 2.28% while the probabiliity that the force is between 14 and 18 kips is 20.37%

Z score is used to determine by how many standard deviations the raw score is above or below the mean. It is given by:

z = (raw score - mean) / standard deviation

Mean value 15.0 (kips) and standard deviation 1.25 (kips).

1) For 17.5:

z = (17.5 - 15) / 1.25 = 2

P(x > 17.4) = 1 - P(z < 2) = 1 - 0.9772 = 0.0228

2) For 14:

z = (14 - 15)/1.25 = 0.8

For 18:

z = (18 - 15)/1.25 = 2.4

P(14 < x < 18) = P(z < 2.4) - P(z < 0.8) = 0.9918 - 0.7881 = 0.2037

The probabiliity that the force is greater than 17.5 kips is 2.28% while the probabiliity that the force is between 14 and 18 kips is 20.37%

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Answer:

a) $430

b) -$90

Step-by-step explanation:

Given a random variable X with possible values

[tex]\large x_1,x_2,...x_n[/tex]

with respective probabilities of occurrence  

[tex]\large P(x_1),P(x_2),...P(x_n)[/tex]

then the expected value E(X) of X is

[tex]\large x_1*P(x_1)+x_2*P(x_2)+...+x_n*P(x_n)[/tex]

a)

The expected collision payment would then be

0*0.85 + 500*0.04 + 1000*0.04 + 3000*0.03 + 5000*0.02 + 8000*0.01 + 10000*0.01 = $430

So the insurance premium that would enable the company to break even is $430

b)

The expected value of the collision policy for a policyholder is the expected payments from the company minus the cost of coverage:

$430 - $520 = -$90

Why does the policyholder purchase a collision policy with this expected value?

Because the policyholder does not know what the probability of having an accident is in her particular case.

Besides, it is better to have the policy and not need it than to need it and not have it.

The probability computed shows that the collision insurance premium that would enable the company to break even is $430.

The collision insurance premium that would enable the company to break even will be calculated thus:

= (0 × 0.85) + (500 × 0.04) + (1000 × 0.04) + (3000 × 0.03) + (5000 × 0.02) + (8000 × 0.01 + (10000 × 0.01)

= $430

The expected value of the collision policy for a policyholder will be:

= $430 - 520

= -$90

In conclusion, the policyholder purchase a collision policy with this expected because he doesn't know the probability of having an accident.

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Answer:

B is 22.12 degrees; ∠C is 57.88°; c=29.24

Step-by-step explanation:

So, first, it's important to draw a diagram of the triangle the problem is talking about (see  attached picture).

Once the triangle has been drawn, we can visualize it better and determine what to do. So first, we are going to find what the value of angle B is by using law of sines:

[tex]\frac{sin B}{b}=\frac{sin A}{a}[/tex]

which can be solved for angle B:

[tex]sin B=b\frac{sin A}{a}[/tex]

[tex] B= sin^{-1}(b\frac{sin A}{a})[/tex]

and substitute the values we already know:

[tex] B= sin^{-1}(13\frac{sin 100 ^{o}}{34})[/tex]

which yields:

B=22.12°

Once we know what the angle of B is, we can now find the value of angle C by using the fact that the sum of the angles of any triangle is equal to 180°. So:

A+B+C=180°

When solving for C we get:

C=180°-A-B

C=180°-22.12°-|00°=57.88°

So once we know what angle C is, we can go ahead and find the length of side c by using the law of sines again:

[tex]\frac{c}{sin C}=\frac{a}{sin A}[/tex]

and solve for c:

[tex]c=sin C \frac{a}{sin A}[/tex]

so we can now substitute for the values we already know:

[tex]c=sin(57.88^{o})\frac{34}{sin(100^{o})}[/tex]

which yields:

c=29.24

Answer:

[tex]F_{T}(t)=\frac{1}{2} +\frac{1}{2}(1-e^{-\lambda t})[/tex]

Step-by-step explanation:

The cumulative distribution function (CDF) F(x),"describes the probability that a random variableX with a given probability distribution will be found at a value less than or equal to x".

The exponential distribution is "the probability distribution of the time between events in a Poisson process (a process in which events occur continuously and independently at a constant average rate). It is a particular case of the gamma distribution". The probability density function is given by:

[tex]P(X=x)=\lambda e^{-\lambda x}[/tex]

We are interested on the cumulative distribution function (CDF) of Jane’s waiting time. Let T the random variable that represent the Jane’s waiting time, and t possible value for the random variable T.

For [tex]t\geq 0[/tex] the CDF for the waiting time is given by:

[tex]F_{T}(t) =P(T\leq t)[/tex]

For this case we can use the total rule of probability and the conidtional probability. We have two options, find 0 or 1 customer ahead of Jane, and for each option we have one possibility in order to find the CDF, like this:

[tex]F_{T}(t)=P(T\leq t|0 customers)P(0 customers)+ P(T\leq t|1 customers)P(1 customers)[/tex]

On this case the probability that we need to wait if we have o customers ahead is [tex]P(T\leq t|0 customers)=1[/tex] since if we don't have a customer ahead, so on this case Jane will not wait. And assuming that the probability of find 0 or 1 customers ahead of Jane is equal we have [tex]P(0 customers)=P(1 customers)=\frac{1}{2}[/tex]. And replacing we have:

[tex]F_{T}(t)=\frac{1}{2} +\frac{1}{2}\int_{0}^{t} \lambda e^{-\lambda \tau} d\tau = \frac{1}{2}+\frac{1}{2}[-e^{-\lambda t}+e^{-\lambda 0}]= \frac{1}{2} +\frac{1}{2}(1-e^{-\lambda t})[/tex]

Jane's waiting time scenario can be split into two cases, when there are no customers ahead of her, and when there is one customer ahead of her. For these two cases, the Cumulative Distribution Function (CDF) of Jane's waiting time can be calculated as [tex]0.5 + 0.5 \times (1 - e^{-\lambda w})[/tex]. This answer is derived using the law of total probability and the definition of the CDF for the exponential distribution.

The question asks for the cumulative distribution function (CDF) of Jane’s waiting time at the bank. This is a probability question related to the exponential distribution and cumulative distribution function. The situation can be represented using two sub-situations: one, when there is no customer ahead (happens with 0.5 probability), Jane has no waiting time, and two, when there is a customer (happens with 0.5 probability) and Jane has to wait for an exponentially distributed time with parameter λ.

The cumulative distribution function (CDF) of an exponential random variable is given by [tex]F(x) = 1 - e^{-\lambda x}[/tex] when x ≥ 0. Using the law of total probability, the cumulative distribution function (CDF) of Jane's waiting time W can be obtained as  [tex]P(W \leq w) = P(0 \text{ customer}) \cdot P(W \leq w | 0 \text{ customer}) + P(1 \text{ customer}) \cdot P(W \leq w | 1 \text{ customer}) = 0.5 \cdot 1 + 0.5 \cdot (1 - e^{-\lambda w})[/tex]when w ≥ 0. Or in other words, Jane's waiting time has a CDF of [tex]0.5 + 0.5 \times (1 - e^{-\lambda w})[/tex]

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Answer:

Step-by-step explanation:

the record for the 100 m dash is 12.45 seconds. Since the length of times in the table is relative to the record time, 12.45 seconds,

Pablo's time - 12.45 = -0.12

Pablo's time = -0.12 + 12.45 = 12.33 seconds.

Cindy's time - 12.45 = 0.34

Cindy's time = 0.34 + 12.45 = 12.79 seconds

Eddie's time - 12.45 = -0.15

Eddie's time = -0.15 + 12.45 = 12.3 seconds.

Sam's time - 12.45 = 0.21

Sam's time = 0.21 + 12.45 = 12.66 seconds.

The following statements are true

A. Pablo and Eddie both broke the school record this year.

C. Sam ran the 100-m dash faster than Cindy.

D. Eddie was the fastest runner in the race.

Answer:

A. Pablo and Eddie both broke the school record this year.

Step-by-step explanation:

Current record is 12.45seconds  

Pablo record is -0.12 seconds which is 12.45 – 0.12 = 12.33 seconds  

Eddie’s record is -0.15 seconds which is 12.45 – 0.15 = 12.30 seconds.  

Answer:

a) 0

b) 0.579 is the probability that chloride concentration is less than 105.

c) 0.32 is the probability that chloride concentration differs from the mean by more than 1 standard deviation.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 104

Standard Deviation, σ = 5

We are given that the distribution of blood chloride concentration is a bell shaped distribution that is a normal distribution.

Formula:

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

a) Since normal distribution s a continuous distribution, the probability for a particular value is zero. Therefore,

[tex]P(x =105) = 0[/tex]

b) P(chloride concentration is less than 105)

[tex]P(x < 105) = P(z < \displaystyle\frac{105-104}{5}) = P(z < 0.2)[/tex]

Calculating the value from the standard normal table we have,

[tex]P(z<0.2) = 0.579 = 57.9\%\\P( x < 105) = 57.9\%[/tex]

c) P(chloride concentration differs from the mean by more than 1 standard deviation)

Since it is a normal distribution, the Empirical rule shows that 68% falls within the first standard deviation, 95% within the first two standard deviations, and 99.7% within the first three standard deviations.

= 1 - P(chloride concentration within the mean by 1 standard deviation)

= 1 - 0.68 = 0.32

Final answer:

The probability of a single value is 0 in a normal distribution. The probability that the chloride concentration is less than 105 mmol/L can be found using the z-score and CDF of the normal distribution. The probability of the concentration being more than one standard deviation from the mean involves calculating the z-scores for one standard deviation away from the mean and using standard normal distribution probabilities.

Explanation:

The question involves the application of probability theory and statistics, specifically the properties of the normal distribution. Here are the answers to the student's questions:

(a) In a normal distribution, the probability of a continuous random variable taking on any single value is technically zero because there are an infinite number of potential values it could take. Therefore, the probability that the chloride concentration equals exactly 105 mmol/L is 0.

(b) To find the probability that the chloride concentration is less than 105 mmol/L, we use the cumulative distribution function (CDF) for a normal distribution. The z-score is calculated by subtracting the mean from the observed value and dividing by the standard deviation: z = (105 - 104) / 5 = 0.20. Using standard normal distribution tables or a calculator, we find the probability corresponding to a z-score of 0.20.

(b) The probability that the chloride concentration differs from the mean by more than 1 standard deviation involves finding the probability of being outside the interval [99, 109], since one standard deviation from the mean is 104 ± 5. We calculate the z-scores for 99 and 109, which are -1 and +1, respectively. Using the normal distribution, we find the probabilities associated with these z-scores and then subtract them from 1 to get the probability of being more than one standard deviation away from the mean.

Answer:

114

Step-by-step explanation:

Given that two intervals  are confidence intervals for μ = mean resonance frequency (in hertz) for all tennis rackets of a certain type.

They are

i) [tex](113.4, 114.6)[/tex] and

ii) [tex](113.1, 114.9)[/tex]

We know that confidence interval has centre as the mean value

Hence we find the average of lower and upper bounds to find out the sample mean.

Sample mean in i) [tex]\frac{113.4+114.6}{2} =114[/tex]

Sample mean in ii) [tex]\frac{113.1+114.9}{2} =114[/tex]

Thus we find that value of sample mean =114

The value of the sample mean resonance frequency is 114 hertz.

The value of the sample mean (in hertz) resonance frequency can be found by taking the average of the lower and upper bounds of the confidence intervals. For the first interval (113.4, 114.6), the sample mean is (113.4 + 114.6) / 2 = 114 hertz. For the second interval (113.1, 114.9), the sample mean is (113.1 + 114.9) / 2 = 114 hertz. So the value of the sample mean resonance frequency is 114 hertz.

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Answer: y=x^2+2

Step-by-step explanation:

y=x^2+2 (answer choice 3) is parabolic and not linear

that is the answer

Final answer:

The requirements for constructing a 95% confidence interval and performing a hypothesis test are met. The 95% confidence interval for the true proportion of teenage smokers is approximately 6.99% to 11.1%, indicating that the proportion may have changed from the previous 15%.

Explanation:

To ascertain if there is sufficient evidence to conclude that the percentage of teenagers who smoke has changed from 15%, we should conduct both a confidence interval analysis and a hypothesis test.

Requirements for Confidence Interval and Hypothesis Testing

Both require a random sample and a sufficiently large sample size to warrant using a normal approximation for the distribution of sample proportions. For a 95% confidence interval and hypothesis testing regarding proportions, the sample should satisfy the conditions np≥15 and n(1-p)≥15 where n is the sample size and p is the hypothesized proportion.

In this case, with a sample size of 785 and the previous proportion of 0.15, the conditions 785(0.15)≥ 15 and 785(0.85)≥ 15 are both met, satisfying the criteria to proceed with the confidence interval and hypothesis test.

95% Confidence Interval for True Proportion

To calculate the 95% confidence interval for the proportion of teenage smokers, use the formula:

Proportion = x/n

Margin of Error (ME) = z*(√[p(1-p)/n])

Lower Bound = Proportion - ME

Upper Bound = Proportion + ME

Where:

x is the number of successes in the sample (71 teenagers acknowledged smoking)

n is the sample size (785)

z is the z-score corresponding to the desired confidence level (approximately 1.96 for 95%)

Substituting in the values we get:

Proportion = 71/785 = 0.0904

ME = 1.96*√[0.0904(1-0.0904)/785] ≈ 0.0205

Lower Bound = 0.0904 - 0.0205 ≈ 0.0699

Upper Bound = 0.0904 + 0.0205 ≈ 0.111

Based on this confidence interval, we have evidence that the proportion of smokers may differ from the previous proportion of 15% as it does not include 0.15 in the interval.

Answer:

  x > 6

Step-by-step explanation:

Subtract $1329 from both sides, then divide by $40.

  $1,329 + $40x > $1,569

  $40x > $240

  x > 6

Marie needs to save for more than 6 more weeks.

_____

Comment on the answer

The given inequality is written using the > symbol. Marie would have exactly enough to cover the projected cost after 6 weeks, but it is appropriate from a finance point of view for her to save more than the required amount. It will take more than 6 weeks for her to do that.

In order to estimate the allowed angle error when measuring the height of a tree, you should first use the given distance and the angle of elevation to calculate the tree's height. Then calculate the potential height range considering the 1.5% error, and figure out the corresponding angles with these heights. The angle error can be estimated by the difference between these two angles.

To solve this problem, one can use the principles of trigonometry. Specifically, the tangent function in trigonometry (tan θ = height / distance) can be used to estimate the height of the tree. Here, the distance from the tree (104 feet) and the angle of elevation (27.5°) are given.

We must first calculate the tree height using tan θ = height / distance. After that, we find that the tree is approximately 58.4 ft tall.

In this case, a 1.5% error in the height measurement would be 0.015 * 58.4 ft = 0.876 feet. Using the equation θ = arctan (height / distance), we can figure out the angle error. Enter the heights 58.4 + 0.876 ft and 58.4 - 0.876 into the equation and use the distance 104104 ft to calculate the corresponding angles. The allowed error for the angle follows as the difference of these two angles.

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To ensure the tree height is within 1.5% accuracy, we calculated that the allowable error in the angle measurement is ±0.22 degrees. This was determined by converting the tolerable height range back to angles using inverse tangent. The primary method used is trigonometry.

To determine the allowed error in the angle measurement for a tree with a height accurate to within 1.5%, we follow these steps:

Given that d = 104 ft and tan(27.5°) ≈ 0.5197, we can calculate: h ≈ 104 × 0.5197 ≈ 54.05 ft

So, 54.05 ft ± 0.015 × 54.05 ft ≈ 54.05 ft ± 0.81 ft

Tree Height thus ranges from approximately 53.24 ft to 54.86 ft.

We convert the bounds of height back to angles using inverse tangent:

The allowed error in the angle measurement is the difference from the original angle:

± |27.5° - 27.28°| ≈ ±0.22° and ± |27.72° - 27.5°| ≈ ±0.22°

Therefore, the allowable error in the angle measurement to ensure a height accuracy within 1.5% is ±0.22 degrees.

Answer:

0.0239

Step-by-step explanation:

A publisher reports that 49% of their readers own a personal computer.

Claim : . A marketing executive wants to test the claim that the percentage is actually different from the reported percentage.

[tex]H_0:p = 0.49\\H_a:p\neq 0.49[/tex]

A random sample of 200 found that 42% of the readers owned a personal computer.

No. of people owned a personal computer = [tex]42\% \times 200[/tex]

                                                                       = [tex]\frac{42}{100} \times 200[/tex]

                                                                       = [tex]84[/tex]

[tex]x = 84\\n = 200[/tex]

We will use one sample proportion test  

[tex]\widehat{p}=\frac{x}{n}[/tex]

[tex]\widehat{p}=\frac{84}{200}[/tex]

[tex]\widehat{p}=0.42[/tex]

Formula of test statistic =[tex]\frac{\widehat{p}-p}{\sqrt{\frac{p(1-p)}{n}}}[/tex]

                                       =[tex]-1.98[/tex]

Now refer the p value from the z table

p value =0.0239

Hence The p value of test statistic is 0.0239

Answer:

narrower, reduced

Step-by-step explanation:

Given that a sample of a given size is used to construct a 95% confidence interval for the population mean with a known population standard deviation

we have confidence interval is margin of error on either side of mean.

Margin of error= Critical value * std error

Std error in turn is inversely proportinal to square root of n

For same confidence level, if sample size is increased we get a less margin of error and hence a narrower confidence interval.

when a confidence interval is narrowed, probability of making a type I error would be reduced.

Hence we get

If a bigger sample had been used instead, then the 95% confidence interval would have been __narrower______ and the probability of making an error would have been ___reduced_____.

A larger sample size would result in a narrower 95% confidence interval and a decreased probability of making an error. This is due to less variability and increased precision with increased sample size.

If a larger sample size had been used to construct a 95% confidence interval for the population mean, the interval would have been narrower. This is because larger sample sizes typically result in less variability. In fact, as the sample size increases, the estimate of the population mean becomes more accurate, and thus the confidence interval around that estimate becomes tighter, containing less of the probability distribution.

The probability of making an error, or in other words, the margin of error, decreases with an increase in sample size. This reduction in the level of uncertainty is due to reduced variability seen with larger samples. By using a larger sample, the chances of the confidence interval not containing the population mean, thus making an error, decrease.

To give an example, if you initially have a 95% confidence interval of (67.02, 68.98) from a smaller sample and you then generate a bigger sample, you could get a confidence interval like (67.18, 68.82). This interval is narrower, representing the reduced variability and increased precision from the larger sample size.

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Answer:

Option b is right

Step-by-step explanation:

Given that  a sociologist is interested in testing if the average age at puberty of girls living is Thailand is different from the average age of puberty of girls living in Japan. To test her theory, she selects a random sample of 50 young Thai women and 50 young Japanese women and determines that the average age at puberty of the women in Japan is 1.7 years younger than the average age at puberty of the Thai women

If average of Japanese women is decreased by 0.8 years, we would get mean difference = [tex]1.7+0.8=2.5[/tex]

Since mean difference increases test statistic would increase thus reducing p value, i.e. chances of rejecting H0 higher.

The sample std devition would not change because of this.

Answer:

Dependent

Step-by-step explanation:

Given that there are 15 balls in an urn. 7 of the balls are red, and 8 of the balls are blue

when we do with replacement each time probability for red or blue remains the same.

a)  the probability that if you draw one ball, it will be blue=[tex]\frac{8}{15}[/tex]

(b) If you draw two balls without replacement,  the probability that you draw a red and then a blue

=[tex]\frac{7}{15} *\frac{8}{14} \\=\frac{4}{15}[/tex]

(c) If you draw two balls with replacement,  the probability that you draw a red and then a blue

=[tex]\frac{7}{15} *\frac{8}{15} \\=\frac{56}{225}[/tex]

(d) If you draw three balls without replacement,  the probability that you draw at least two red balls

=P(2 red, 1 blue)+P(3 blue)

=[tex]\frac{7C2*8C1}{15C3} +\frac{7C3}{15C3} \\=\frac{203}{455} =\frac{29}{65}[/tex]

(e) If you draw three balls with replacement,  the probability that you draw at least two red balls

= P(X≥2) where x is binomial with n=3 and p = 7/15

= 0.4506

(f) We play a game where I draw a ball first without replacement, and then you draw one. I draw a blue ball, and then you draw a red ball.

No when first ball is drawn without replacement the next probability for blue and red would be changed.  So dependent.

The probability of various scenarios involving drawing balls from an urn is calculated, including with and without replacement. The events of drawing a blue ball and a red ball after each other are examined in terms of their independence.

In order to answer the given question, we need to calculate the probability for each scenario:

(a) The probability of drawing a blue ball with replacement is

8/15

.

(b) The probability of drawing a red and then a blue ball without replacement is

(7/15) * (8/14)

.

(c) The probability of drawing a red and then a blue ball with replacement is

(7/15) * (8/15)

.

(d) The probability of drawing at least two red balls without replacement can be calculated by finding the probability of drawing two red balls and the probability of drawing three red balls, and then adding them together:

(7/15) * (6/14) + (7/15) * (6/14) * (5/13)

.

(e) The probability of drawing at least two red balls with replacement can be calculated by finding the probability of drawing two red balls and the probability of drawing three red balls, and then adding them together:

(7/15) * (7/15) + (7/15) * (7/15) * (7/15)

(f) No, the events of you drawing a blue ball and the student drawing a red ball after that are

not independent

because drawing the blue ball affects the probability of drawing a red ball for the student.

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Answer:

a) P(X=41)= 0.0086 = 0.86%

b) P(36≤X≤38)== 0.1215 = 12.15%

c) would be unusual that fewer that 20 murders were cleared because P(X≤20) = 0.000081 =0.0081%

Step-by-step explanation:

if the random variable X= number of cleared murders , then X follows a binomial distribution:

P(X=x)= n!/[(n-x)!*x!]*p^x*(1-p)^(n-x)

where P(X=x)= probability of x cleared murders, n= number of murders selected=50 , p= probability for a murder to be cleared (67%)

therefore

a) P(X=41)=50!/[(50-41)!*41!]*0.67^41*0.33^(50-41)= 0.0086

b) P(36≤X≤38)= P(X≤38) - P(X≤36) = 0.9371- 0.8156 = 0.1215

c) P(X≤20) = 0.000081

therefore would be unusual that fewer that 20 murders were cleared

Final answer:

To construct the 95% confidence interval for the average time spent in a particular job at General Motors, we calculated the margin of error and then added and subtracted it from the sample mean, resulting in a confidence interval of approximately (6.1331, 6.8669).

Explanation:

To construct a 95% confidence interval for the average time spent in a particular job at General Motors, we utilize the sample mean, standard deviation, and the sample size to compute the margin of error and the confidence interval. Given that the sample mean is 6.5 years and the standard deviation is 1.7 years with a sample size of 85, we can calculate the confidence interval as follows:

This confidence interval suggests that we are 95% confident that the true average time spent in the job before promotion at General Motors lies between approximately 6.1 and 6.9 years.

Answer:

Step-by-step explanation:

The solid is a cuboid with length of 6 inches, a width of 3 inches, and a height of 2 inches.

The volume is l× w×h = 6×3×2 = 36inches^3

Therefore, the following statements are true about the Solid

1) The volume of the solid is 36 in.3.

2) The perimeter of one of its faces is 10 inches.(2width + 2 height = 2×2 + 2×3 = 10 inches)

3) The area of one of its faces is 6 inches^2. (width × height = 3×2 = 6 inches)

4) The area of one of its faces is 12 in.2.(length × height = 6×2 = 12 inches^2)

Answer : The correct options are:

The perimeter of one of its faces is 10 inches.

The area of one of its faces is 6 inch².

The area of one of its faces is 12 inch².

The volume of the solid is 36 inches³.

Step-by-step explanation :

Formula used for perimeter of one face (rectangle) is:

Perimeter = 2(l+b)

Perimeter = 2(b+h)

Perimeter = 2(l+h)

Formula used for area of one face (rectangle) is:

Area = l × b

Area = b × h

Area = l × h

Formula used for volume of solid (cuboid) is:

volume of solid = l × b × h

Given:

l = length = 6 inches

b = width = 3 inches

h = height = 2 inches

Now we have to calculate the perimeter of one face.

Perimeter = 2(l+b)

Perimeter = 2(6+3) = 18 inch

Perimeter = 2(b+h)

Perimeter = 2(3+2) = 10 inch

Perimeter = 2(l+h)

Perimeter = 2(6+2) = 16 inch

Now we have to calculate the area of one face.

Area = l × b

Area = 6 × 3 = 18 inch²

Area = b × h

Area = 3 × 2= 6 inch²

Area = l × h

Area = 6 × 2 = 12 inch²

Now we have to calculate the volume of solid.

volume of solid = l × b × h

volume of solid = 6 × 3 × 2

volume of solid = 36 inches³

The volume of solid is, 36 inches³

Answer:

0.4545 or 45%

Step-by-step explanation:

Do-Gooders toys for boys = 5

Southside toys for boys = 6

Given that the first toy pulled out of the bin is a toy for boys, to find the probability that this toy was donated by the Do-Gooders, the number of toys for girls donate shouldn't be considered since there's a zero-chance that the toy in question is for girls. Therefore:

[tex]Total\ sample \ space = 5+6 =11\\P(Do-Gooders)=\frac{5}{11} =0.4545[/tex]

There's a 0.4545 or 45% probability that it was donated by the Do-Gooders.

Answer:

a) [tex]\hat \mu =\bar X=24.694[/tex]

b) The 99% confidence interval would be given by (24.409;24.979)

c) Since the 99% confidence interval not contains the exposed to ambient temperatures of 24 degrees Celsius, and since the animals "that can maintain a body temperature different from the surroundings would be considered evidence of this regulating capability", on this case we can conclude that the sea crabs are able to regulate their body temperature.

Step-by-step explanation:

1) 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".

Data: 24.33, 24.61, 24.67, 24.64, 24.42, 24.97, 25.23, 24.73, 24.90, 24.44

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

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

s represent the sample standard deviation

n=10 represent the sample size  

Part a

A point of estimate for the population mean is the sample mean, given by this formula:

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

If we apply this formula we got that [tex]\hat \mu =\bar X=24.694[/tex]

Part b

In order to calculate the confidence interval first we need to calculate the sample deviation given by this formula:

[tex]s=\frac{\sum_{i=1}^n (x_i -\bar x)}{n-1}[/tex]

If we use this formula we got that [tex]s=0.277[/tex]

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

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

In order to calculate the critical value [tex]t_{\alpha/2}[/tex] we need to find first the degrees of freedom, given by:

[tex]df=n-1=10-1=9[/tex]

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: "=-T.INV(0.005,9)".And we see that [tex]t_{\alpha/2}=3.25[/tex]

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

[tex]24.694-3.25\frac{0.277}{\sqrt{10}}=24.409[/tex]    

[tex]24.694+3.25\frac{0.277}{\sqrt{10}}=24.979[/tex]

So on this case the 99% confidence interval would be given by (24.409;24.979)    

Part c

Since the 99% confidence interval not contains the exposed to ambient temperatures of 24 degrees Celsius, and since the animals "that can maintain a body temperature different from the surroundings would be considered evidence of this regulating capability", on this case we can conclude that the sea crabs are able to regulate their body temperature.

Answer:[tex]\frac{1}{\mu }+\frac{1}{\lambda }[/tex]

Step-by-step explanation:

If a machine is replaced at some time t, then the expected time until next failure is [tex]\frac{1}{\mu }[/tex]

and the time between the checks is exponentially distributed with rate \lambda, the expected time until next failure is [tex]\frac{1}{\lambda }[/tex]  

Because of memory less property of the exponential  The answer is

[tex]\frac{1}{\mu }+\frac{1}{\lambda }[/tex]

Answer:

The correct option is D) [tex]f'(t)-g'(t) > 0[/tex]

Step-by-step explanation:

Consider the provided information.

People are entering a building at a rate modeled by f (t) people per hour and exiting the building at a rate modeled by g (t) people per hour,

The change of number of people in building is:

[tex]h(x)=f(t)-g(t)[/tex]

Where f(t) is people entering in building and g(t) is exiting from the building.

It is given that "The functions f and g are non negative and differentiable for all times t."

We need to find the the rate of change of the number of people in the building.

Differentiate the above function with respect to time:

[tex]h'(x)=\frac{d}{dt}[f(t)-g(t)][/tex]

[tex]h'(x)=f'(t)-g'(t)[/tex]

It is given that the rate of change of the number of people in the building is increasing at time t.

That means [tex]h'(x)>0[/tex]

Therefore, [tex]f'(t)-g'(t)>0[/tex]

Hence, the correct option is D) [tex]f'(t)-g'(t) > 0[/tex]

The rate of change of the number of people in the building is increasing at time t and with the help of this statement the correct option is D).

Given :

The change of the number of people in the building is given by:

[tex]h(x) = f(t) - g(t)[/tex]

To determine the inequality, differentiate the above equation with respect to time.

[tex]h'(x)=f'(t)-g'(t)[/tex]

Now, it is given that the rate of change of the number of people in the building is increasing at time t. That means:

[tex]h'(x)>0[/tex]

[tex]f'(t)-g'(t)>0[/tex]

Therefore, the correct option is D).

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Answer: [tex]\mu_x=18\text{ hours}[/tex]

[tex]\sigma_x=4\text{ hours}[/tex]

Step-by-step explanation:

We know that mean and standard deviation of sampling distribution is given by :-

[tex]\mu_x=\mu[/tex]

[tex]\sigma_x=\dfrac{\sigma}{\sqrt{n}}[/tex]

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

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

n= sample size .

In the given situation, we have

[tex]\mu=18\text{ hours}[/tex]

[tex]\sigma=6\text{ hours}[/tex]

n= 2

Then, the expected mean and the standard deviation of the sampling distribution will be :_

[tex]\mu_x=\mu=18\text{ hours}[/tex]

[tex]\sigma_x=\dfrac{\sigma}{\sqrt{n}}=\dfrac{6}{\sqrt{2}}=4.24264068712\approx4[/tex]  [Rounded to the nearest whole number]

Hence, the the expected mean and the standard deviation of the sampling distribution :

[tex]\mu_x=18\text{ hours}[/tex]

[tex]\sigma_x=4\text{ hours}[/tex]