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Natali5045456 [20]

2 years ago

11

Assume that the histograms below represent the amount of time it took two different groups of 100 people to run at a 5k race. As

sume that the mean of each histogram is the same. Which group has a greater mean absolute deviation. Why?

1 answer:

Effectus [21]2 years ago

6 0

I have know idea I would just guess if I were you

Explanation

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Please help with these math questions

SSSSS [86.1K]

1) a
2) d
3) 7 and 2
4) a and d
5) 50

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2 years ago

Find two numbers that multiply to give u 12 and add to give you 13

Shkiper50 [21]

12 and 1
12+1 = 13
12*1= 12

6 0

2 years ago

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The mean cost of a five pound bag of shrimp is 40 dollars with a standard deviation of 7 dollars. If a sample of 51 bags of shri

damaskus [11]

Answer:

$P(|\bar{x}-40| > 0.6)=0.5404$

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 40 dollars

Standard Deviation, σ = 7 dollars

Sample size,n = 51

We are given that the distribution of cost of shrimp is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

Standard error due to sampling =

$=\dfrac{\sigma}{\sqrt{n}} = \dfrac{7}{\sqrt{51}} = 0.9801$

P(sample mean would differ by true mean by more than 0.6)

$P(|\bar{x}-40| > 0.6)\\\Rightarrow = 1-P(39.4$

$P(39.4 \leq x \leq 40.6) \\\\= P(\displaystyle\frac{39.4 - 40}{0.9801} \leq z \leq \displaystyle\frac{40.6-40}{0.9801})\\\\ = P(-0.6121 \leq z \leq 0.6121)\\= P(z \leq 0.6121) - P(z < -0.6121)\\= 0.7298 - 0.2702 =0.4596$

$P(|\bar{x}|-40 > 0.6)\\= 1-0.4596=0.5404$

0.5404 is the required probability.

6 0

2 years ago

a total of 321 tickets were sold for the school play. They were either adult tickets or student tickets. The number of student t

Aleks [24]

<span>Joey, divide the total of 321 by 3. that's the number of adult tickets. double that answer and that's the number of student tickets.</span>

5 0

2 years ago

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A pollster would like to estimate the fraction p of people in a population who intend to vote for a particular candidate. How la

Lera25 [3.4K]

Answer:

$n=\frac{0.5(1-0.5)}{(\frac{0.02}{1.96})^2}=2401$

And rounded up we have that n=2401

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".

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

Solution to the problem

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ . And the critical value would be given by:

$z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96$

The margin of error for the proportion interval is given by this formula:

$ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$ (a)

And on this case we have that $ME =\pm 0.02$ and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}$ (b)

The estimated proportion is 0.5 for this case since we don't have prior information. And replacing into equation (b) the values from part a we got:

$n=\frac{0.5(1-0.5)}{(\frac{0.02}{1.96})^2}=2401$

And rounded up we have that n=2401

5 0

3 years ago