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Sergeeva-Olga [200]
2 years ago
9

Shania's test scores in 8 subjects were 88, 91, 85, 74, 69, 72, 80, and 87. Which measure should Shania use to find how much the

scores on her tests vary, on average, from the mean score?
A.
mode
B.
mean absolute deviation
C.
interquartile range
D.
mean
Mathematics
2 answers:
dalvyx [7]2 years ago
4 0
The answer is B, Mean Absolute Deviation.
poizon [28]2 years ago
3 0

✧・゚: *✧・゚:*    *:・゚✧*:・゚✧

                  Hello!

✧・゚: *✧・゚:*    *:・゚✧*:・゚✧

❖ The correct answer would be B. Mean Absolute Deviation.

~ ʜᴏᴘᴇ ᴛʜɪꜱ ʜᴇʟᴘꜱ! :) ♡

~ ᴄʟᴏᴜᴛᴀɴꜱᴡᴇʀꜱ

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Let X represent the amount of gasoline (gallons) purchased by a randomly selected customer at a gas station. Suppose that the me
Alexus [3.1K]

Answer:

a) 18.94% probability that the sample mean amount purchased is at least 12 gallons

b) 81.06% probability that the total amount of gasoline purchased is at most 600 gallons.

c) The approximate value of the 95th percentile for the total amount purchased by 50 randomly selected customers is 621.5 gallons.

Step-by-step explanation:

To solve this question, we use 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 normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For sums, we can apply the theorem, with mean \mu and standard deviation s = \sqrt{n}*\sigma

In this problem, we have that:

\mu = 11.5, \sigma = 4

a. In a sample of 50 randomly selected customers, what is the approximate probability that the sample mean amount purchased is at least 12 gallons?

Here we have n = 50, s = \frac{4}{\sqrt{50}} = 0.5657

This probability is 1 subtracted by the pvalue of Z when X = 12.

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

By the Central Limit theorem

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

Z = \frac{12 - 11.5}{0.5657}

Z = 0.88

Z = 0.88 has a pvalue of 0.8106.

1 - 0.8106 = 0.1894

18.94% probability that the sample mean amount purchased is at least 12 gallons

b. In a sample of 50 randomly selected customers, what is the approximate probability that the total amount of gasoline purchased is at most 600 gallons.

For sums, so mu = 50*11.5 = 575, s = \sqrt{50}*4 = 28.28

This probability is the pvalue of Z when X = 600. So

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

Z = \frac{600 - 575}{28.28}

Z = 0.88

Z = 0.88 has a pvalue of 0.8106.

81.06% probability that the total amount of gasoline purchased is at most 600 gallons.

c. What is the approximate value of the 95th percentile for the total amount purchased by 50 randomly selected customers.

This is X when Z has a pvalue of 0.95. So it is X when Z = 1.645.

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

1.645 = \frac{X- 575}{28.28}

X - 575 = 28.28*1.645

X = 621.5

The approximate value of the 95th percentile for the total amount purchased by 50 randomly selected customers is 621.5 gallons.

5 0
2 years ago
ASAP PLZZZ HELP
lisabon 2012 [21]

Answer:

group one c d f h group two a g e b

Step-by-step explanation:

Hope this helps

5 0
3 years ago
A student wants to report on the number of meals his friends buy each week. The collected data are below: 4 19 3 3 2 3 2 4 Which
Naddik [55]

Answer:

B

Step-by-step explanation:

7 0
2 years ago
A bag contains 4 blue marbles and 3 red marbles. What is the probability of picking a blue marble, putting it back in the bag, t
patriot [66]

Answer:

3/14

Step-by-step explanation:

P=3/(3+4+7)=3/14 because each of 14 marbles is equally likely to be drawn.

Hope this helped :)

Have a great day!

8 0
2 years ago
Whats the mean, media, maximum,minimum,mode and range of 1,2,1,3,4
ludmilkaskok [199]
First put the numbers in order least to greatest. 
(1,1,2,3,4)
The mean is the average of the group of numbers.
Too find it you add all the numbers up and divide it by how many numbers there are.
In this case it would be 11 divided by 5 so the mean would be 2 \frac{1}{5}

Median is just another word for middle. After putting the numbers in order, the number in the center of the group is the median, in this case it would be 2.

The maximum would just be the largest number which is self explantory (4)

The minimum would be the lowest number (1)

I was taught to see mode as the word "most". All you do is see which number occurs the most. If all the numbers only occur once, your answer would be "No mode"


<span>It is that simple! The Range is the difference between the lowest and highest number. So all you have to do is subtract the maximum by the minimum. In this case it would be 4-1 = 3
</span>
3 0
2 years ago
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