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3 years ago

19.) Here's the data (sorted) of the ages of 91 women who won the Oscar for Best Actress in a Leading Role:

Mathematics

1 answer:

exis [7]3 years ago

6 0

Step-by-step explanation:

The five number summary includes the mean, the median, the mode, the range, and the interquartile range.

1) Mean = The sum of all observations / number of observations.

Mean = (21 + 22 + 22 + 24 + 24 + 24 + 24 + 25 + 26 + 26 + 26 + 26 + 26 + 26 + 26 + 26 + 26 + 27 + 27 + 27 + 27 + 28 + 28 + 28 + 28 + 29 + 29 + 29 + 29 + 29 + 30 + 30 + 30 + 30 + 30 + 30 + 31 + 31 + 31 + 32 + 32 + 33 + 33 + 33 + 33 + 33 + 33 + 33 + 34 + 34 + 34 + 34 + 34 + 35 + 35 + 35 + 35 + 35 + 37 + 37 + 37 + 37 + 38 + 38 + 38 + 39 + 39 + 39 + 41 + 41 + 41 + 41 + 42 + 42 + 44 + 45 + 45 + 45 + 47 + 49 + 49 + 54 + 60 + 60 + 61 + 61 + 61 + 62 + 62 + 74 + 81)/91.

Mean = 3285/91

Mean = 36 (round to the nearest whole number).

2) Median = The central point of the data.

91 divided by 2 gives 45.5, which means 45th number will be the median of this data set. So, Median = 33.

3) Mode = The mostly repeated observation of the data.

Mode = 26 (repeated 9 times).

4) Range = Largest observation - Smallest observation

Range = 81 - 21 = 60.

5) Interquartile Range = Upper Quartile - Lower Quartile.

Upper Quartile is the data point at the 75% position of the data.

Lower Quartile is the data point at the 25% position of the data.

Upper Quartile position = 91*0.75 = 68.25 (68th position).

So that observation is 39.

Lower Quartile position = 91*0.25 = 22.75 (22nd position).

So that observation is 28.

Therefore, interquartile range = 39 - 28 = 11!!!

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The cost of a week long vacation in a city has a mean of 1000 and a variance of 1200. The city imposes a tax which will raise al

matrenka [14]

Answer:

The coefficient of variation after the tax is imposed is 0.033

Step-by-step explanation:

Given

$\mu =1000$ --- mean

$\sigma^2 = 1200$ --- variance

$tax = 10\%$

Required

The coefficient of variation

The coefficient of variation is calculated using:

$CV = \frac{\sqrt{\sigma^2}}{\mu}$

After the tax, the new mean is:

$\mu_{new} = \mu * (1 + tax)$

$\mu_{new} = 1000 * (1 + 10\%)$

$\mu_{new} = 1100$

And the new variance is:

$\sigma^2_{new} = \sigma^2 * (1 + tax)$

$\sigma^2_{new} = 1200 * (1 + 10\%)$

$\sigma^2_{new} = 1320$

So, we have:

$CV = \frac{\sqrt{\sigma^2}}{\mu}$

$CV = \frac{\sqrt{1320}}{1100}$

$CV = \frac{36.33}{1100}$

$CV = 0.033$

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3 years ago

HELP ME PLEASE IM BEGGING

agasfer [191]

95 is the area
Hope this helps :p

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2 years ago

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What is X if Y is 12.8 and Z is 16 in the right triangle below?

Natalija [7]

Answer:

Im 90% sure its H.

Step-by-step explanation:

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6/30=3/15 true or false

Dennis_Churaev [7]

Answer: True

Step-by-step explanation: This is because they are equivalent fractions

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2 years ago

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After once again losing a football game to the college'ss arch rival, the alumni association conducted a survey to see if alumni

antoniya [11.8K]

Answer:

a) The 90% confidence interval would be given (0.561;0.719).

b) $p_v =P(z>2.8)=1-P(z$

c) Using the significance level assumed $\alpha=0.01$ we see that $p_v$ so we have enough evidence at this significance level to reject the null hypothesis. And on this case makes sense the claim that the proportion is higher than 0.5 or 50%.

Step-by-step explanation:

1) Data given and notation

n=100 represent the random sample taken

X=64 represent were in favor of firing the coach

$\hat p=\frac{64}{100}=0.64$ estimated proportion for were in favor of firing the coach

$p_o=0.5$ is the value that we want to test since the problem says majority

$\alpha$ represent the significance level (no given, but is assumed)

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)

p= population proportion of Americans for were in favor of firing the coach

Part a

The confidence interval would be given by this formula

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

For the 90% confidence interval the value of $\alpha=1-0.9=0.1$ and $\alpha/2=0.05$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=1.64$

And replacing into the confidence interval formula we got:

$0.64 - 1.64 \sqrt{\frac{0.64(1-0.64)}{100}}=0.561$

$0.64 + 1.64 \sqrt{\frac{0.64(1-0.64)}{100}}=0.719$

And the 90% confidence interval would be given (0.561;0.719).

Part b

We need to conduct a hypothesis in order to test the claim that the proportion exceeds 50%(Majority). :

Null Hypothesis: $p \leq 0.5$

Alternative Hypothesis: $p >0.5$

We assume that the proportion follows a normal distribution.

This is a one tail upper test for the proportion of union membership.

The One-Sample Proportion Test is "used to assess whether a population proportion $\hat p$ is significantly (different,higher or less) from a hypothesized value $p_o$ ".

Check for the assumptions that he sample must satisfy in order to apply the test

a)The random sample needs to be representative: On this case the problem no mention about it but we can assume it.

b) The sample needs to be large enough

$np_o =100*0.64=64>10$

$n(1-p_o)=100*(1-0.64)=36>10$

Calculate the statistic

The statistic is calculated with the following formula:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o(1-p_o)}{n}}}$

On this case the value of $p_o=0.5$ is the value that we are testing and n = 100.

$z=\frac{0.64 -0.5}{\sqrt{\frac{0.5(1-0.5)}{100}}}=2.8$

The p value for the test would be:

$p_v =P(z>2.8)=1-P(z$

Part c

Using the significance level assumed $\alpha=0.01$ we see that $p_v$ so we have enough evidence at this significance level to reject the null hypothesis. And on this case makes sense the claim that the proportion is higher than 0.5 or 50%.

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3 years ago