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lilavasa [31]
3 years ago
6

The owner of the car dealership decides to treat the value 22 as an outlier.which measure of center or spread is affected the mo

st if the owner removes this outlier ?
Mathematics
1 answer:
erastova [34]3 years ago
4 0

Answer:

See explanation

Step-by-step explanation:

Solution:-

- The effect of an outlier on the mean, median and range is to be investigated.

- Mean: It is the average of all the values. If the outlier "22" is lies on the upper spectrum of the center value. If the outlier is removed the value of center or mean will decrease.

- Median: The median value is mostly defined as the value around which their is a cluster of data. The value of the outlier "22" if close to that cluster of data points is omitted there will be small deviation in the value of median. If the value of the outlier "22" if far away to that cluster of data points is omitted there will be significant deviation in the value of median.

- Range: Is defined by the uppermost and lowermost value from a set of data points that is considered. The value of outlier will equally effect either of these limits depending where the outlier lies close to upper limit or lower limit of the range.

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8. Solve for y:<br> 24 = 3y
Gnom [1K]

Answer:

y=8

Step-by-step explanation:

24/3=8

3*8=24

hope this helps :3

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3 years ago
Milk has a pH of 6.7, which is slightly acidic. Cheesemakers add culture to milk to lower the pH, making it more acidic and turn
vladimir1956 [14]

Answer:

e. The probability of observing a sample mean of 5.11 or less, or of 5.29 or more, is 0.018 if the true mean is 5.2.

Step-by-step explanation:

We have a two-tailed one sample t-test.

The null hypothesis claims that the pH is not significantly different from 5.2.

The alternative hypothesis is that the mean pH is significantly different from 5.2.

The sample mean pH is 5.11, with a sample size of n=50.

The P-value of the test is 0.018.

This P-value corresponds to the probability of observing a sample mean of 5.11 or less, given that the population is defined by the null hypothesis (mean=5.2).

As this test is two-tailed, it also includes the probability of the other tail. That is the probability of observing a sample with mean 5.29 or more (0.09 or more from the population mean).

Then, we can say that, if the true mean is 5.2, there is a probability P=0.018 of observing a sample of size n=50 with a sample mean with a difference bigger than 0.09 from the population mean of the null hypothesis (5.11 or less or 5.29 or more).

The right answer is e.

7 0
3 years ago
A tobacco company claims that the amount of nicotene in its cigarettes is a random variable with mean 2.2 and standard deviation
Aleksandr-060686 [28]

Answer:

0% probability that the sample mean would have been as high or higher than 3.1 if the company’s claims were true.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central limit theorem

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 2.2, \sigma = 0.3, n = 100, s = \frac{0.3}{\sqrt{100}} = 0.03

What is the approximate probability that the sample mean would have been as high or higher than 3.1 if the company’s claims were true?

This is 1 subtracted by the pvalue of Z when X = 3.1. So

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{3.1 - 2.2}{0.03}

Z = 30

Z = 30 has a pvalue of 1.

1 - 1 = 0

0% probability that the sample mean would have been as high or higher than 3.1 if the company’s claims were true.

4 0
4 years ago
Covert 15/7/8 to a decimal
serg [7]

Answer:

0.268

Step-by-step explanation:

In the attached file

4 0
3 years ago
Read 2 more answers
You are biking to the park. After six minutes 40% of the distance still need to be traveled. How long does it take to bike to th
nikklg [1K]

Answer:

y = (100-10x)\%

It takes <em>10 minutes</em> to the park.

Step-by-step explanation:

Given that:

Time taken = 6 minutes

Distance left to be traveled = 40%

To find:

The equation to represent the distance y, that still needs to be traveled after x minutes.

Solution:

As per given statement, distance traveled in 6 minutes = 60%

We can say that:

Distance traveled each minute = \frac{60\%}{6} = 10%

After 1 minute, distance traveled = 10%

Distance left = 100 - 10 = 90%

After 2 minutes, distance traveled = 20%

Distance left = 100 - 20 = 80%

After x minutes, distance traveled = 10x%

Distance left, y = (100 - 10x)%

Therefore, the equation is:

y = (100-10x)\%

Time taken by the bike to reach to the park can be found by putting y = 0 (i.e. distance left to be traveled = 0)

0 = 100 -10x\\\Rightarrow 10x = 100\\\Rightarrow x = \bold{10\ minutes}

8 0
3 years ago
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