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garri49 [273]
2 years ago
13

Consider the following dot plot. Of the following statements, which two characteristics of this dot plot make the median a bette

r choice than the mean to summarize the center of the distribution?
The data are skewed and there is an outlier.
The data are symmetric and there is an outlier.
The peak is equal to the median and the data are skewed.
The mean is equal to the median and the data are symmetric.

Mathematics
2 answers:
Readme [11.4K]2 years ago
6 0
The data is skewed and there is an outlier.
SashulF [63]2 years ago
3 0

A box and whisker plot is used to display information about the range, the median and the quartiles .The box plot  is a standardized way of displaying the distribution of data based on the five number summary: minimum, first quartile, median, third quartile, and maximum.Data can be "skewed", meaning it tends to have a long tail on one side or the other .

THere are points in the graph which are falling outside the data set We have outlier in the set of datas.

Among the given options the option that holds true for the given graph is

The data are skewed and there is an outlier.



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2.A production process manufactures items with weights that are normally distributed with mean 10 pounds and standard deviation
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Answer:

Step-by-step explanation:

Given that:

population mean = 10

standard deviation = 0.1

sample mean = 9.8 < x > 10.2

The z score can be computed as:

z = \dfrac{\bar x - \mu}{\sigma}

if x > 10.2

z = \dfrac{10.2- 10}{0.1}

z = \dfrac{0.2}{0.1}

z = 2

If x < 9.8

z = \dfrac{9.8- 10}{0.1}

z = \dfrac{-0.2}{0.1}

z = -2

The p-value = P (z ≤ 2) + P (z ≥ 2)

The p-value = P (z ≤ 2) + ( 1 -  P (z ≥ 2)

p-value = 0.022750 +(1 -   0.97725)

p-value = 0.022750 +  0.022750

p-value = 0.0455

Therefore; the probability of defectives  = 4.55%

the probability of acceptable = 1 - the probability of defectives

the probability of acceptable = 1 - 0.0455

the probability of acceptable = 0.9545

the probability of acceptable = 95.45%

4.55% are defective or 95.45% is acceptable.

sampling distribution of proportions:

sample size n=1000

p = 0.0455

The z - score for this distribution at most 5% of the items is;

z = \dfrac{0.05 - 0.0455}{\sqrt{\dfrac{0.0455\times 0.9545}{1000}}}

z = \dfrac{0.0045}{\sqrt{\dfrac{0.04342975}{1000}}}

z = \dfrac{0.0045}{\sqrt{4.342975 \times 10^{-5}}}

z = 0.6828

The p-value = P(z ≤ 0.6828)

From the z tables

p-value = 0.7526

Thus, the probability that at most 5% of the items in a given batch will be defective = 0.7526

The z - score for this distribution for at least 85% of the items is;

z = \dfrac{0.85 - 0.9545}{\sqrt{\dfrac{0.0455\times 0.9545}{1000}}}

z = \dfrac{-0.1045}{\sqrt{\dfrac{0.04342975}{1000}}}

z = −15.86

p-value = P(z ≥  -15.86)

p-value = 1 - P(z <  -15.86)

p-value = 1 - 0

p-value = 1

Thus, the probability that at least 85% of these items in a given batch will be acceptable = 1

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