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ad-work [718]
3 years ago
15

The data represented by the graph is normally distributed and adheres to the 68-95-99.7 rule. the standard deviation of this dat

a is (drop down menu), and (drop down menu2)% of the population will fall below the mark of 7.
drop down menu1-
-6.6
1.8
2.2
4.8

drop down menu2-
34
68
84
95

Mathematics
1 answer:
jenyasd209 [6]3 years ago
8 0

Answer:

i)

the standard deviation of this data is 2.2

ii) 84%

Step-by-step explanation:

We have been informed that the data represented by the graph is normally distributed and adheres to the 68-95-99.7 rule. This implies that we shall be dealing with the applications of the empirical rule.

In a normal distribution, the mean is also the mode as well as the median of the distribution. This implies that the mean is at the center of the distribution. In our case, the mean is 4.8. The vertical lines represents data points that lie 1, 2 and 3 standard deviations away from the mean.

From the graph, both 11.4 and -1.8 lie 3 standard deviations away from the mean on either side. We can thus use the following equation to solve for the unknown standard deviation;

mean + 3(sigma) = 11.4

where sigma is the standard deviation.

our mean is 4.8, therefore;

4.8 + 3(sigma) = 11.4

3(sigma) = 6.6

sigma = 2.2

ii)

In order to determine the percentage of the population that will fall below the mark of 7, we shall first determine how many standard deviations away from the mean the value 7 lies.

To do this, we can use the following equation;

4.8 + n(2.2) = 7

2.2n = 2.2

n = 1

Therefore, the data point 7 lies one standard deviation away from the mean to the right.

The percentage of the population that will fall below one standard deviation away from the mean is;

68 + 13.5 + 2.35 = 83.85%

Using the empirical rule

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2 years ago
According to the February 2008 Federal Trade Commission report on consumer fraud and identity theft, 23% of all complaints in 20
Murljashka [212]

Answer:

There is not enough evidence to support the claim that Alaska had a lower proportion of identity theft than 23%.  

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 1432

p = 23% = 0.23

Alpha, α = 0.05

Number of theft complaints , x = 321

First, we design the null and the alternate hypothesis  

H_{0}: p = 0.23\\H_A: p < 0.23

This is a one-tailed test.  

Formula:

\hat{p} = \dfrac{x}{n} = \dfrac{321}{1432} =0.2241

z = \dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}

Putting the values, we get,

z = \displaystyle\frac{0.2241-0.23}{\sqrt{\frac{0.23(1-0.23)}{1432}}} = -0.5305

Now, we calculate the p-value from the table.

P-value = 0.298

Since the p-value is greater than the significance level, we fail to reject the null hypothesis and accept the null hypothesis.

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