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sertanlavr [38]
2 years ago
8

True or​ false: when comparing two​ populations, the larger the standard​ deviation, the more dispersion the distribution​ has,

provided that the variable of interest from the two populations has the same unit of measure. choose the correct answer below.
a. ​false, because the larger the standard deviation​ is, the less dispersion the distribution has.
b. ​true, because the standard deviation is the difference between the largest and smallest observation. when the standard deviation is​ larger, there is more distance between the largest and smallest​ observation, and​ therefore, more dispersion in the distribution.
c. ​true, because the standard deviation describes how​ far, on​ average, each observation is from the typical value. a larger standard deviation means that observations are more distant from the typical​ value, and​ therefore, more dispersed.
d. ​false, because the standard deviation measures the spread of the​ distribution, not the dispersion of the distribution.
Mathematics
1 answer:
Nataliya [291]2 years ago
7 0
<span>It would be...
C) true, because the standard deviation describes how​ far, on​ average, each observation is from the typical value. a larger standard deviation means that observations are more distant from the typical​ value, and​ therefore, more dispersed.
</span>
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Assume the returns from an asset are normally distributed. The average annual return for the asset is 17.4 percent and the stand
Jlenok [28]

Answer:

0.13%

Step-by-step explanation:

Mean, u = 17.4%

standard deviation, sd = 27.5%

We are looking for the probability that the money will double in value. Therefore we are looking for the value of x = 100% (meaning the value is 100% more that the previous year)

Find the value of z-score

z = (x-u)/sd

  = (100 - 17.4) / 27.5 = 3.003

Find the z score using the z-table

From z-table at P(z<3.003) = 0.9987

So, for the probability that it the money will be double is

P (Z>2.8982) = 1 - P(z<2.8982)

                      = 1- 0.9987

                      = 0.0013  = 0.13%

6 0
2 years ago
Patrick and Brooklyn are making decisions about their bank accounts. Patrick wants to deposit $300 as a principle amount, with a
irina [24]
For this we will use formula that is letting us to input: interest rate, starting funds, how often intereset rate is implemented, period we are observing. Formula looks like this:

M = S(1+ \frac{i}{cp})^cp*y

where M is money, S is starting funds, "i" is interest rate, cp is compounding period and y is number of years. now we express and calculated for both of them and get

M = 318,479 for Patricks investement.

M = 331,482 for Brooklyn.

Which means Brooklyn's method will pay of more.
8 0
2 years ago
Read 2 more answers
In March 2015, the Public Policy Institute of California (PPIC) surveyed 7525 likely voters living in California. In the survey,
balandron [24]

Answer:

(\hat p_A -\hat p_B) \pm z_{\alpha/2} \sqrt{\frac{\hat p_A(1-\hat p_A)}{n_A} +\frac{\hat p_B (1-\hat p_B)}{n_B}}

And for this case the confidence interval is given by: (0.275, 0.305)

We can estimate the proportion difference as:

\hat p_D = \frac{0.275+0.305}{2}=0.29

And the margin of error would be:

ME=0.305-0.29=0.015

So then for this case the possibl two options are:

We are 95% confident that the true difference in proportion of Latinos who view global warming as a serious problem and whites who view global warming as a serious problem is 27.5% to 30.5%.

We are 95% confident that the difference between Latino and white opinions about the severity of global warming is 29% with a margin of error of 1.5%.

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".  

p_A represent the real population proportion of Latinos who view global warming as a serious problem

\hat p_A=0.75 represent the estimated proportion Latinos who view global warming as a serious problem

n_A is the sample size required of Latinos who view global warming as a serious problem

p_B represent the real population proportion of white who view global warming as a serious problem

\hat p_B =0.46 represent the estimated proportion  of whitewho view global warming as a serious problem

n_B is the sample size required of white

z represent the critical value for the margin of error  

Solution to the problem

The population proportion have the following distribution  

p \sim N(p,\sqrt{\frac{p(1-p)}{n}})  

The confidence interval for the difference of two proportions would be given by this formula  

(\hat p_A -\hat p_B) \pm z_{\alpha/2} \sqrt{\frac{\hat p_A(1-\hat p_A)}{n_A} +\frac{\hat p_B (1-\hat p_B)}{n_B}}

And for this case the confidence interval is given by: (0.275, 0.305)

We can estimate the proportion difference as:

\hat p_D = \frac{0.275+0.305}{2}=0.29

And the margin of error would be:

ME=0.305-0.29=0.015

So then for this case the possibl two options are:

We are 95% confident that the true difference in proportion of Latinos who view global warming as a serious problem and whites who view global warming as a serious problem is 27.5% to 30.5%.

We are 95% confident that the difference between Latino and white opinions about the severity of global warming is 29% with a margin of error of 1.5%.

4 0
2 years ago
A jar of marbles contains 7 blue marbles, 12 yellow marbles, 8 green marbles, and 5 red marbles. Use this information to answer
Andreyy89
The correct answer is 9/32
6 0
2 years ago
The average score of all golfers for a particular course has a mean of 64 and a standard deviation of 3. Suppose 36 golfers play
kondaur [170]

Answer:

the probability that the exceeded 65 = 0.3707

The average score of the 36 golfers exceeded 65

                        = 36 X 0.3707 = 13.3452

Step-by-step explanation:

<u>Step 1</u>:-

The average score of all golfers for a particular course has a mean of 64 and a standard deviation of 3.

mean (μ) = 64

standard deviation (σ) =3

by using normal distribution

given (μ) = 64 and  (σ) =3

i) when x =65

z = \frac{x-mean}{S.D} = \frac{65-64}{3} = 0.33 >0

P( X≥ 65) = P(z≥0.33)

              = 0.5 - A(z₁)

              = 0.5 - 0.1293 (see normal table)

             = 0.3707

The average score of the 36 golfers exceeded 65

                        = 36 X 0.3707 = 13.3452

                                               

6 0
2 years ago
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