Answer:
(a) P([tex]\bar X[/tex] [tex]\geq[/tex] 76) = 0.2327
(b) P(73 < [tex]\bar X[/tex] < 75) = 0.5035
(c) P([tex]\bar X[/tex] < 74.8) = 0.77035
Step-by-step explanation:
We are given that the mean of a population is 74 and the standard deviation is 16.
Assuming the data follows normal distribution.
Let [tex]\bar X[/tex] = sample mean
The z-score probability distribution for sample mean is given by;
Z = [tex]\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] ~ N(0,1)
where, [tex]\mu[/tex] = population mean = 74
[tex]\sigma[/tex] = standard deviation = 16
n = sample size
The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.
(a) Probability that a random sample of size 34 yielding a sample mean of 76 or more is given by = P([tex]\bar X[/tex] [tex]\geq[/tex] 76)
P([tex]\bar X[/tex] [tex]\geq[/tex] 76) = P( [tex]\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] [tex]\geq[/tex] [tex]\frac{76-74}{\frac{16}{\sqrt{34} } }[/tex] ) = P(Z [tex]\geq[/tex] 0.73) = 1 - P(Z < 0.73)
= 1 - 0.7673 = 0.2327
The above probability is calculated by looking at the value of x = 0.73 in the z table which has an area of 0.7673.
(b) Probability that a random sample of size 120 yielding a sample mean of between 73 and 75 is given by = P(73 < [tex]\bar X[/tex] < 75) = P([tex]\bar X[/tex] < 75) - P([tex]\bar X[/tex] [tex]\leq[/tex] 73)
P([tex]\bar X[/tex] < 75) = P( [tex]\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] < [tex]\frac{75-74}{\frac{16}{\sqrt{120} } }[/tex] ) = P(Z < 0.68) = 0.75175
P([tex]\bar X[/tex] [tex]\leq[/tex] 73) = P( [tex]\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] [tex]\leq[/tex] [tex]\frac{73-74}{\frac{16}{\sqrt{120} } }[/tex] ) = P(Z [tex]\leq[/tex] -0.68) =1 - P(Z < 0.68)
= 1 - 0.75175 = 0.24825
Therefore, P(73 < [tex]\bar X[/tex] < 75) = 0.75175 - 0.24825 = 0.5035
The above probability is calculated by looking at the value of x = 0.68 in the z table which has an area of 0.75175.
(c) Probability that a random sample of size 218 yielding a sample mean of less than 74.8 is given by = P([tex]\bar X[/tex] < 74.8)
P([tex]\bar X[/tex] < 74.8) = P( [tex]\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] < [tex]\frac{74.8-74}{\frac{16}{\sqrt{218} } }[/tex] ) = P(Z < 0.74) = 0.77035
The above probability is calculated by looking at the value of x = 0.74 in the z table which has an area of 0.77035.
In a study of the accuracy of fast food drive-through orders, one restaurant had 36 orders that were not accurate among 324 orders observed. Use a 0.01 significance level to test the claim that the rate of inaccurate orders is equal to 10%. Does the accuracy rate appear to be acceptable?
Identify the null and alternative hypotheses for this test. Choose the correct answer below.
Answer:
We do not have sufficient evidence to reject the claim that ,the rate of inaccurate orders is equal to 10%.
Step-by-step explanation:
We want to use a 0.01 significance level to test the claim that the rate of inaccurate orders is equal to 10%.
We set up our hypothesis to get:
[tex]H_0:p=0.10[/tex]------->null hypothesis
[tex]H_1:p\ne0.10[/tex]------>alternate hypothesis
This means that: [tex]p_0=0.10[/tex]
Also, we have that, one restaurant had 36 orders that were not accurate among 324 orders observed.
This implies that: [tex]\hat p=\frac{36}{324}=0.11[/tex]
The test statistics is given by:
[tex]z=\frac{\hat p-p_0}{\sqrt{\frac{p_0(1-p_0)}{n} } }[/tex]
We substitute to obtain:
[tex]z=\frac{0.11-0.1}{\sqrt{\frac{0.1(1-0.1)}{324} } }[/tex]
This simplifies to:
[tex]z=0.6[/tex]
We need to calculate our p-value.
P(z>0.6)=0.2743
Since this is a two tailed test, we multiply the probability by:
The p-value is 2(0.2723)=0.5486
Since the significance level is less than the p-value, we fail to reject the null hypothesis.
We do not have sufficient evidence to reject the claim that ,the rate of inaccurate orders is equal to 10%.
Which of the following illustrates the product rule for logarithmic equations?
1.log2 (4x) = log2(4) / log2(x)
2.log2 (4x) = log2(4)x log2(x)
3.log2 (4x) = log2(4)-log2 (x)
4.log2 (4x) = log2(4)+log2 (x)
Answer:log2 (4x) = log2(4)+log2 (x) it is D
Step-by-step explanation:
Gabriel finds some wooden boards in the backyard with lengths of 5 feet, 2.5 feet and 4 feet. He decides he wants to make a triangular garden in the yard and uses the triangle inequality rule to see if it will work.
PLEASE ANSWER IMMEDIATELY Which sums prove that the boards will create a triangular outline for the garden? Select all that apply.
5 + 2.5 > 4
5 + 2.5 < 4
4 + 2.5 > 5
4 + 2.5 < 5
4 + 5 > 2.5
first one to answer will get brainliest
Answer:
Third one
4 + 2.5 > 5
Step-by-step explanation:
To form a triangle, the longest side should be less than the sum of the two shorter sides (triangular inequality)
Longest side < sum of the shorter ones
Longest side: 5
Shorter ones: 2.5 & 4
5 < 2.5 + 4
This is the same as:
4 + 2.5 > 5
Answer:
1 3 5
Step-by-step explanation:
Bankers at a large financial institution created the linear regression model dˆ=0.37−0.0004s to predict the proportion of customers who would default on their loans, d , based on the customer’s credit score, s . For a customer with a credit score of 700, which of the following is true?
Answer:
The correct option is (a).
Step-by-step explanation:
The complete question is:
Bankers at a large financial institution created the linear regression model dˆ=0.37−0.0004s to predict the proportion of customers who would default on their loans, d , based on the customer’s credit score, s . For a customer with a credit score of 700, which of the following is true?
(a) The default proportion is predicted to be 0.09.
(b) The default proportion will be 0.09.
(c) The default proportion is predicted to be approximately 1.75 million.
(d) The default proportion will be approximately 1.75 million.
(e) The default proportion is predicted to be 0.28.
Solution:
The linear regression model is used to predict the value of the response or dependent variable based on only one explanatory or independent variable.
The general form of a linear regression model is:
[tex]\hat y=\alpha +\beta x[/tex]
Here,
y = dependent variable
x = independent variable
α = intercept
β = slope
The linear regression model to predict the proportion of customers who would default on their loans, based on the customer’s credit score is:
[tex]\hat d=0.37-0.0004\ s[/tex]
d = default on loans
s = customer’s credit score
Compute the predicted value of d for s = 700 as follows:
[tex]\hat d=0.37-0.0004\ s[/tex]
[tex]=0.37-(0.0004\times 700)\\=0.37-0.28\\=0.09[/tex]
Thus, for a customer with a credit score of 700, the default proportion is predicted to be 0.09.
Thus, the correct option is (a).
A researcher collected data of systolic blood pressure and weight for 5 patients, as are shown in the table below.
Patient Systolic Blood Pressure (in mmHg) Weight (in lbs)
1 145 210
2 155 245
3 160 260
4 156 230
5 150 219
1. Draw a scatter plot of systolic blood pressure (response) versus weight (regressor).2. What is the direction of the association?
Answer:
We can see the details in the pic.
Step-by-step explanation:
We can see the details in the pic shown.
A researcher claims that the mean annual cost of raising a child (age 2 and under) by husband-wife families in the U.S. is $13,960. In a random sample of husband-wife families in the U.S. the mean annual cost of raising a child (age 2 and under) is $13,725. The sample consists of 500 children and the population standard deviation is $2,345. At the α = 0.10, is there enough evidence to reject the claim? Use the p-value approach.
Answer:
[tex]z=\frac{13725-13960}{\frac{2345}{\sqrt{500}}}=-2.24[/tex]
[tex]p_v =2*P(z<-2.24)=0.0251[/tex]
If we compare the p value and the significance level given [tex]\alpha=0.1[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is different from 13960 at 10% of signficance.
Step-by-step explanation:
Data given and notation
[tex]\bar X=13275[/tex] represent the sample mean
[tex]\sigma=2345[/tex] represent the sample standard deviation
[tex]n=500[/tex] sample size
[tex]\mu_o =68[/tex] represent the value that we want to test
[tex]\alpha=0.1[/tex] represent the significance level for the hypothesis test.
t would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value for the test (variable of interest)
State the null and alternative hypotheses.
We need to conduct a hypothesis in order to check if the true mean is 13960, the system of hypothesis would be:
Null hypothesis:[tex]\mu = 13690[/tex]
Alternative hypothesis:[tex]\mu \neq 13690[/tex]
If we analyze the size for the sample is > 30 and we know the population deviation so is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:
[tex]z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}[/tex] (1)
z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".
Calculate the statistic
We can replace in formula (1) the info given like this:
[tex]z=\frac{13725-13960}{\frac{2345}{\sqrt{500}}}=-2.24[/tex]
P-value
Since is a two sided test the p value would be:
[tex]p_v =2*P(z<-2.24)=0.0251[/tex]
Conclusion
If we compare the p value and the significance level given [tex]\alpha=0.1[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is different from 13960 at 10% of signficance.
If tan (k•90)=0 then k is an even integer true of false
Answer:
True
Step-by-step explanation:
Use the unit circle
tan(x) = 0 only when x is 0 or a multiple of 180 (in degrees)
Answer:
true
Step-by-step explanation:
A jogger runs along a straight track. The jogger’s position is given by the function p(t), where t is measured in minutes since the start of the run. During the first minute of the run, the jogger’s acceleration is proportional to the square root of the time since the start of the run. Write a differential equation that describes this relationship, where k is a positive constant?
Answer:
[tex]\frac{d^{2}p}{dt^{2}}=k*\sqrt{t}[/tex]
Step-by-step explanation:
Given
The jogger’s position: p(t)
We can express the acceleration a as follows
[tex]a=\frac{d^{2}p}{dt^{2}}[/tex]
then
[tex]\frac{d^{2}p}{dt^{2}}=k*\sqrt{t}[/tex]
only if
[tex]0 min\leq t\leq 1 min[/tex]
The required differential equation will be [tex]\dfrac{d^2P(t)}{dt^2}=k\sqrt t[/tex] or [tex]\dfrac{d^2P(t)}{dt^2}-k\sqrt t=0[/tex].
Given information:
A jogger runs along a straight track. The jogger’s position is given by the function p(t), where t is measured in minutes since the start of the run.
During the first minute of the run, the jogger’s acceleration is proportional to the square root of the time.
Let a be the acceleration of the jogger.
So, the expression for acceleration can be written as,
[tex]a=\dfrac{d}{dt}(\dfrac{dP(t)}{dt})\\a=\dfrac{d^2P(t)}{dt^2}[/tex]
Now, the acceleration is proportional to square root of time.
So,
[tex]a\propto \sqrt t\\a=k\sqrt t\\\dfrac{d^2P(t)}{dt^2}=k\sqrt t[/tex]
Therefore, the required differential equation will be [tex]\dfrac{d^2P(t)}{dt^2}=k\sqrt t[/tex] or [tex]\dfrac{d^2P(t)}{dt^2}-k\sqrt t=0[/tex].
For more details about differential equations, refer to the link:
https://brainly.com/question/1164377
I need help please?!!
Answer:
No, the expression is not linear because the highest power of x is 2.
Answer:
No it isn't it is Quadratic
A group of 2n people, consisting of n men and n women, are to be independently distributed among m rooms. Each woman chooses room j with probability pj while each man chooses it with probability qj,j=1,…,m. Let X denote the number of rooms that will contain exactly one man and one woman. (a) Find µ = E[X] (b) Bound P{|X − µ} > b} for b > 0
Step-by-step explanation:
Assume that
[tex]X_i = \left \{ {{1, If , Ith, room, has,exactly, 1,man, and , 1,woman } \atop {0, othewise} \right.[/tex]
hence,
[tex]x = x_1 + x_2+....+x_m[/tex]
now,
[tex]E(x) = E(x_1+x_2+---+x_m)\\\\E(x)=E(x_1)+E(x_2)+---+E(x_m)[/tex]
attached below is the complete solution
The Institute of Management Accountants (IMA) conducted a survey of senior finance professionals to gauge members’ thoughts on global warming and its impact on their companies. The survey found that 65% of senior professionals that global warming is having a significant impact on the environment. Suppose that you select a sample of 100 senior finance professionals.
1. What is the probability that the sample percentage indicating global warming is having a significant impact on the environment will be between 64% and 69%?
2. The probability is 90% that the sample percentage will be contained within what symmetrical limits of the population percentage?
3. The probability is 95% that the sample percentage will be contained within what symmetrical limits of the population percentage?
Answer:
(1) The probability that the sample percentage indicating global warming is having a significant impact on the environment will be between 64% and 69% is 0.3674.
(2) The two population percentages that will contain the sample percentage with probability 90% are 0.57 and 0.73.
(3) The two population percentages that will contain the sample percentage with probability 95% are 0.55 and 0.75.
Step-by-step explanation:
Let X = number of senior professionals who thought that global warming is having a significant impact on the environment.
The random variable X follows a Binomial distribution with parameters n = 100 and p = 0.65.
But the sample selected is too large and the probability of success is close to 0.50.
So a Normal approximation to binomial can be applied to approximate the distribution of p if the following conditions are satisfied:
np ≥ 10 n(1 - p) ≥ 10Check the conditions as follows:
[tex]np= 100\times 0.65=65>10\\n(1-p)=100\times (1-0.65)=35>10[/tex]
Thus, a Normal approximation to binomial can be applied.
So, [tex]\hat p\sim N(p, \frac{p(1-p)}{n})=N(0.65, 0.002275)[/tex].
(1)
Compute the value of [tex]P(0.64<\hat p<0.69)[/tex] as follows:
[tex]P(0.64<\hat p<0.69)=P(\frac{0.64-0.65}{\sqrt{0.002275}}<\frac{\hat p-p}{\sqrt{\frac{p(1-p)}{n}}}<\frac{0.69-0.65}{\sqrt{0.002275}})[/tex]
[tex]=P(-0.20<Z<0.80)\\=P(Z<0.80)-P(Z<-0.20)\\=0.78814-0.42074\\=0.3674[/tex]
Thus, the probability that the sample percentage indicating global warming is having a significant impact on the environment will be between 64% and 69% is 0.3674.
(2)
Let [tex]p_{1}[/tex] and [tex]p_{2}[/tex] be the two population percentages that will contain the sample percentage with probability 90%.
That is,
[tex]P(p_{1}<\hat p<p_{2})=0.90[/tex]
Then,
[tex]P(p_{1}<\hat p<p_{2})=0.90[/tex]
[tex]P(\frac{p_{1}-p}{\sqrt{\frac{p(1-p)}{n}}}<\frac{\hat p-p}{\sqrt{\frac{p(1-p)}{n}}}<\frac{p_{2}-p}{\sqrt{\frac{p(1-p)}{n}}})=0.90[/tex]
[tex]P(-z<Z<z)=0.90\\P(Z<z)-[1-P(Z<z)]=0.90\\2P(Z<z)-1=0.90\\2P(Z<z)=1.90\\P(Z<z)=0.95[/tex]
The value of z for P (Z < z) = 0.95 is
z = 1.65.
Compute the value of [tex]p_{1}[/tex] and [tex]p_{2}[/tex] as follows:
[tex]-z=\frac{p_{1}-p}{\sqrt{\frac{p(1-p)}{n}}}\\-1.65=\frac{p_{1}-0.65}{\sqrt{\frac{0.65(1-0.65)}{100}}}\\p_{1}=0.65-(1.65\times 0.05)\\p_{1}=0.5675\\p_{1}\approx0.57[/tex] [tex]z=\frac{p_{2}-p}{\sqrt{\frac{p(1-p)}{n}}}\\1.65=\frac{p_{2}-0.65}{\sqrt{\frac{0.65(1-0.65)}{100}}}\\p_{2}=0.65+(1.65\times 0.05)\\p_{1}=0.7325\\p_{1}\approx0.73[/tex]
Thus, the two population percentages that will contain the sample percentage with probability 90% are 0.57 and 0.73.
(3)
Let [tex]p_{1}[/tex] and [tex]p_{2}[/tex] be the two population percentages that will contain the sample percentage with probability 95%.
That is,
[tex]P(p_{1}<\hat p<p_{2})=0.95[/tex]
Then,
[tex]P(p_{1}<\hat p<p_{2})=0.95[/tex]
[tex]P(\frac{p_{1}-p}{\sqrt{\frac{p(1-p)}{n}}}<\frac{\hat p-p}{\sqrt{\frac{p(1-p)}{n}}}<\frac{p_{2}-p}{\sqrt{\frac{p(1-p)}{n}}})=0.95[/tex]
[tex]P(-z<Z<z)=0.95\\P(Z<z)-[1-P(Z<z)]=0.95\\2P(Z<z)-1=0.95\\2P(Z<z)=1.95\\P(Z<z)=0.975[/tex]
The value of z for P (Z < z) = 0.975 is
z = 1.96.
Compute the value of [tex]p_{1}[/tex] and [tex]p_{2}[/tex] as follows:
[tex]-z=\frac{p_{1}-p}{\sqrt{\frac{p(1-p)}{n}}}\\-1.96=\frac{p_{1}-0.65}{\sqrt{\frac{0.65(1-0.65)}{100}}}\\p_{1}=0.65-(1.96\times 0.05)\\p_{1}=0.552\\p_{1}\approx0.55[/tex] [tex]z=\frac{p_{2}-p}{\sqrt{\frac{p(1-p)}{n}}}\\1.96=\frac{p_{2}-0.65}{\sqrt{\frac{0.65(1-0.65)}{100}}}\\p_{2}=0.65+(1.96\times 0.05)\\p_{1}=0.748\\p_{1}\approx0.75[/tex]
Thus, the two population percentages that will contain the sample percentage with probability 95% are 0.55 and 0.75.
The questions concern calculating the probability of a sample proportion falling within a given range and constructing confidence intervals around the population proportion. These questions are related to statistical concepts such as the normal approximation to the binomial distribution and the use of z-scores for interval estimation, which are typically covered in college-level statistics courses.
Explanation:The student has asked about determining the probability that a sample percentage will fall within certain ranges, given a known proportion from a survey conducted by the Institute of Management Accountants (IMA) regarding the impact of global warming on their companies. Specifically, the student is looking to estimate probabilities related to the sample proportion and construct confidence intervals around a population percentage. These are statistical concepts typically covered in college-level courses in probability and statistics, specifically in chapters related to sampling distributions and confidence interval estimation.
To answer the first question, we would need to use the normal approximation to the binomial distribution, since the sample size is large (n=100). The sample proportion p = 0.65, and we can calculate the standard error for the sampling distribution of the sample proportion. However, since the full calculations are not provided here, a specific numerical answer cannot be given.
For the second and third questions, constructing confidence intervals at 90% and 95% requires using the standard error and the appropriate z-scores that correspond to these confidence levels. Again, the specific limits are not calculated here, but the process involves multiplying the standard error by the z-score and adding and subtracting this product from the sample percentage.
Where are the negative rational numbers placed on a vertical number line?
Answer:
on the top the negative rational numbers are placed
Answer:
On a horizontal number line, the positive rational numbers are placed To the right of, To the left of, above or below zero, and the negative rational numbers are placed To the right of, To the left of, above or below zero.
Write an inequality for each statement.
1. Lacrosse practice will not be more than 45 minutes.
2. Mario measures more than 60 inches in height
Write an inequality for each statement.
1. Lacrosse practice will not be more than 45 minutes.
2. Mario measures more than 60 inches in height
3. More than 8000 fans attended the first football game of the Wizard amd Arrowhead Stadium in Kansas City, Missouri. Write an inequality to describe attendance.
3. More than 8000 fans attended the first football game of the Wizard amd Arrowhead Stadium in Kansas City, Missouri. Write an inequality to describe attendance.
Answer:
1. [tex]l \leq 45[/tex] ( in this inequality, the time can be less than or equal to 45, but no more than 45)
2. [tex]m > 60[/tex] (Mario's height is more than 60.)
3. [tex]f > 8000[/tex] (More than 8000 fans attended.)
please help :wildlife society wants to investigate the effects of a chemical spill on the growth of various types of fish. Scientists will start by measuring the lengths and weights of a sample of fish in the area. Which of the following samples should be selected for this study?
A.
fish in local lakes
B.
fish in oceans worldwide
C.
fish at the local zoo
D.
pet fish owned by local citizens
Suppose Julio is a veterinarian who is doing research into the weight of domestic cats in his city. He collects information on 188 cats and finds the mean weight for cats in his sample is 10.97 lb with a standard deviation of 4.41 lb. What is the estimate of the standard error of the mean (SE)
Answer:
The standard error of the mean is 0.3216
Step-by-step explanation:
We are given the following in the question:
Sample size, n = 188
Sample mean =
[tex]\bar{x}= 10.97\text{ lb}[/tex]
Sample standard deviation =
[tex]s = 4.41\text{ lb}[/tex]
We have to estimate the standard error of the mean.
Formula for standard error:
[tex]S.E = \dfrac{s}{\sqrt{n}}[/tex]
Putting values, we get,
[tex]S.E =\dfrac{4.41}{\sqrt{188}} = 0.3216[/tex]
Thus, the standard error of the mean is 0.3216
Evaluate the following expression. Round your answer to two decimal places.
Answer: 2.18
Step-by-step explanation:
log(11)/log(3)=2.18
Answer:
2.18
Step-by-step explanation:
I just did it and got it right
additive inverse of −11/ −13
Answer: The additive inverse of -11/-13 is 11/13
Step-by-step explanation: The additive inverse is the opposite sign. (positive or negative). It's the amount that needs to be added or subtracted to get 0.
Bryce reads in the latest issue of Pigskin Roundup that the average number of rushing yards per game by NCAA Division II starting running backs is 50 with a standard deviation of 8 yards. If the number of yards per game (X) is normally distributed, what is the probability that a randomly selected running back has 64 or fewer rushing yards
Answer:
0.9599 is the probability that a randomly selected running back has 64 or fewer rushing yards.
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 50
Standard Deviation, σ = 8
We are given that the distribution of number of rushing yards per game is a bell shaped distribution that is a normal distribution.
Formula:
[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]
P(running back has 64 or fewer rushing yards)
[tex]P( x \leq 64) = P( z \leq \displaystyle\frac{64 - 50}{8}) = P(z \leq 1.75)[/tex]
Calculation the value from standard normal z table, we have,
[tex]P(x \leq 64) = 0.9599[/tex]
0.9599 is the probability that a randomly selected running back has 64 or fewer rushing yards.
A noted psychic was tested for extrasensory perception. The psychic was presented with 2 0 0 cards face down and asked to determine if each card were one of five symbols: a star, a cross, a circle, a square, or three wavy lines. The psychic was correct in 5 0 cases. Let p represent the probability that the psychic correctly identifies the symbol on the card in a random trial. Assume the 2 0 0 trials can be treated as a simple random sample from the population of all guesses the psychic would make in his lifetime. What do we know about the value of the P -value for the hypothesis test: Ha: p>0.20 ? ( Note: Use the large-sample z statistic. ) a. P-value < 0.01 0.02 < b. P-value < 0.03 0.03 < c. P-value < 0.04 0.05 < d. P-value < 0.10
Answer:
C. P- value < 0.04 0.05
Step-by-step explanation:
hello,
we were given the sample size, n = 200
also the probability that the psychic correctly identifies the symbol on the 200 card is
[tex]p=\frac{50}{200}= 0.25[/tex]
using the large sample Z- statistic, we have
[tex]Z=\frac{p- 0.20}{\sqrt{0.2(1-0.2)/200} }[/tex]
= [tex]\frac{0.25-0.20}{\sqrt{0.16/200}}[/tex]
= 1.7678
thus the P - value for the hypothesis test is P(Z > 1.7678) = 0.039.
from the above, we conclude that the P- value < 0.04, 0.05
Identify if there is a relationship between the variables.
No, there is no relationship because the points are
all up and down.
Yes, the relationship displayed shows arm span
increasing as height increases.
O Yes, the relationship displayed shows arm span
decreasing as height increases.
Answer:
increases
Step-by-step explanation:
Answer:
the answer is:B
Step-by-step explanation:
Lucille has a collection of more than 500 songs on her phone that have a mean duration of 215 seconds and a standard deviation of 35 seconds. Suppose that every week she makes a playlist by taking an SRS of 49 of these songs, and we calculate the sample mean duration ë of the songs in each sample. Calculate the mean and standard deviation of the sampling distribution of ________. seconds L = seconds
Answer:
The mean of the sampling distribution is of 215 seconds and the standard deviation is 5.
Step-by-step explanation:
The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
All songs
Mean 215 seconds, standard deviation 35 seconds
Sample
49
Mean 215, standard deviation [tex]s = \frac{35}{\sqrt{49}} = 5[/tex]
The mean of the sampling distribution is of 215 seconds and the standard deviation is 5.
Answer:
The sample mean would be:
[tex]\mu_{\bar X} = 215 seconds[/tex]
And the deviation:
[tex]\sigma_{\bar X} = \frac{35}{\sqrt{49}}= 5 seconds[/tex]
Step-by-step explanation:
Previous concepts
The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".
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".
Solution to the problem
We know the following info for the random variable X who represent the duration
[tex]\mu = 215, \sigma=35[/tex]
For this case we select a sampel size of n =49>30. So we can apply the central limit theorem. From the central limit theorem we know that the distribution for the sample mean [tex]\bar X[/tex] is given by:
[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]
The sample mean would be:
[tex]\mu_{\bar X} = 215 seconds[/tex]
And the deviation:
[tex]\sigma_{\bar X} = \frac{35}{\sqrt{49}}= 5 seconds[/tex]
please help, im so confused!
Which point is on the graph of the function
f(x) = One-half(2)x?
(0, 1)
(0, 2)
(1, One-half)
(1, 1)
Answer:
D (1,1)
Step-by-step explanation:
Answer:
D. (1,1)
Hope it works!
Step-by-step explanation:
find the value of x in 2(x+1)=4
Answer:
x = 1
Step-by-step explanation:
We need to solve for x by isolating the variable.
First, expand the parentheses:
2(x + 1) = 4
2 * x + 2 * 1 = 4
2x + 2 = 4
Then subtract by 2:
2x + 2 - 2 = 4 - 2
2x = 2
Finally, divide by 2:
2x/2 = 2/2
x = 1
Thus, x = 1.
Hope this helps!
Answer:
x = 1
Step-by-step explanation:
Wellllll....
X = the number to solve the equation to equal 4
2(x +1) = 4
Multiply 2 by x and 1 by x
2x +2 = 4
Solve!
2x = 4-2
2x = 2
x = 1
Hope this works!
Rosa
A movie theater is keeping data on the number of tickets sold and the price of a single ticket. Which scatterplot is
correctly labeled?
Answer: 3rd option
Step-by-step explanation:
Answer:the third option is the answer
Step-by-step explanation:I just took the assignment on edge
A number that is 12 times greater than -4
Answer:
My answer is not correct, do not read
i need to find the length
Answer: I am pretty sure it is 7
Step-by-step explanation:
the smaller one is to smaller than the bigger one so you would just add two to the 5
Answer:
7.5
Step-by-step explanation:
The triangles are proportional, so 6/4=x/5
1.5=x/5
x=1.5*5=7.5
NEED HELP ASAP!! FILL IN THE BLANKS!! I'LL MARK YOU BRAINLIEST IF YOU ANSWER RIGHT!!!
Answer:
(x+8)^2 = 127
Step-by-step explanation:
R=1/6 the problem is 3r2
Step-by-step explanation:
Putting the value of r
3 ( 1/6) 2
So 3 × (1/6) = 0.5 × 2 = 1
2) 5, 28, 16, 32,5, 16, 48, 29, 5, 35
Mean:?
Median:?
Mode:?
Range:?
The mean is 21.9, the median is 22, the mode is 5 and the range is 43.
Important information:
The given data values are 5, 28, 16, 32,5, 16, 48, 29, 5, 35.Mean, Median, Mode, Range:Mean of the data set is:
[tex]Mean=\dfrac{5+28+16+32+5+16+48+29+5+35}{10}[/tex]
[tex]Mean=\dfrac{219}{10}[/tex]
[tex]Mean=21.9[/tex]
Arrange the data set in asccending order.
5, 5, 5, 16, 16, 28, 29, 32, 35, 48
The number of observation is 10, which is an even number. So, the median is average of [tex]\dfrac{10}{2}=5th[/tex] term and [tex]\dfrac{10}{2}+1=6th[/tex].
[tex]Median=\dfrac{16+28}{2}[/tex]
[tex]Median=\dfrac{44}{2}[/tex]
[tex]Median=22[/tex]
Mode is the most frequent value.
In the given data set 5 has the highest frequency 3. So, the mode of the data is 5.
Range is the data set is:
[tex]Range=Maximum-Minimum[/tex]
[tex]Range=48-5[/tex]
[tex]Range=43[/tex]
Therefore, the mean is 21.9, the median is 22, the mode is 5 and the range is 43.
Find out more about 'Mean, Median, Mode, Range' here:
https://brainly.com/question/361615
In 2011, a U.S. Census report determined that 71% of college students work. A researcher thinks this percentage has changed since then. A survey of 110 college students reported that 91 of them work. Is there evidence to support the reasearcher's claim at the 1% significance level? A normal probability plot indicates that the population is normally distributed.
a) Determine the null and alternative hypotheses.
H0: p =
Ha:P Select an answer (Put in the correct symbol and value)
b) Determine the test statistic. Round to two decimals.
c) Find the p-value. Round to 4 decimals.
P-value =
Answer:
(a) Null Hypothesis, [tex]H_0[/tex] : p = 71%
Alternate Hypothesis, [tex]H_A[/tex] : p [tex]\neq[/tex] 71%
(b) The test statistics is 3.25.
(c) The p-value is 0.0006.
Step-by-step explanation:
We are given that a U.S. Census report determined that 71% of college students work. A researcher thinks this percentage has changed since then.
A survey of 110 college students reported that 91 of them work.
Let p = proportion of college students who work
(a) Null Hypothesis, [tex]H_0[/tex] : p = 71% {means that % of college students who work is same as 71% since 2011}
Alternate Hypothesis, [tex]H_A[/tex] : p [tex]\neq[/tex] 71% {means that % of college students who work is different from 71% since 2011}
The test statistics that will be used here is One-sample z proportion statistics;
T.S. = [tex]\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] ~ N(0,1)
where, [tex]\hat p[/tex] = sample proportion of college students who reported they work = [tex]\frac{91}{110}[/tex] = 82.73%
n = sample of students = 110
(b) So, test statistics = [tex]\frac{\frac{91}{110}-0.71}{\sqrt{\frac{\frac{91}{110}(1-\frac{91}{110})}{110} } }[/tex]
= 3.25
The test statistics is 3.25.
(c) P-value of the test statistics is given by the following formula;
P-value = P(Z > 3.25) = 1 - P(Z [tex]\leq[/tex] 3.25)
= 1 - 0.99942 = 0.0006
So, the p-value is 0.0006.
The null and alternative hypotheses is [tex]\rm H_0:[/tex] p = 71% and [tex]\rm H_a[/tex] : p [tex]\neq[/tex] 71%, the value test statistics is 3.25, and the p-value is 0.0006.and this can be determined by using the given data.
Given :
In 2011, a U.S. Census report determined that 71% of college students work.A survey of 110 college students reported that 91 of them work.a) The hypothesis is given by:
Null hypothesis -- [tex]\rm H_0:[/tex] p = 71%
Alternate Hypothesis -- [tex]\rm H_a[/tex] : p [tex]\neq[/tex] 71%
b) The statistics test is given by:
[tex]\rm TS = \dfrac{\hat{p}-p}{\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}} }[/tex]
[tex]\rm TS = \dfrac{\dfrac{91}{110}-0.71}{\sqrt{\dfrac{\dfrac{91}{110}(1-\dfrac{91}{110})}{110}} }[/tex]
Simplify the above expression.
TS = 3.25
c) The p-value is given by:
P-value = P(Z>3.25) = 1 - P(Z [tex]\leq[/tex] 3.25)
= 1 - 0.99942
= 0.0006
For more information, refer to the link given below:
https://brainly.com/question/17716064