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3 years ago

Mean, Median, Mode and Range. Easy hard math. Please show work and give answers . Need help appreciate you all

Mathematics

1 answer:

tigry1 [53]3 years ago

4 0

Answer:

Step-by-step explanation:

quiz scores:

a. arrange them all from least to greatest to figure out the median first

your quiz: 15, 16, 17, 18, 19 middle # = 17 median = 17

friend's quiz: 12, 13, 17, 20, 20 middle # = 17 median = 17

now for mean add all of your quiz scores up and divide by how many #s

15 + 16 + 17 + 18 + 19 = 85 / 5 = 17 mean = 17

now your friend's:

12 + 13 + 17 + 20 + 20 = 82 / 5 = 16.4 mean = 16.4

mode is which number is used the most

your mode = you don't have a mode your friend's mode = 20

b. you have a higher mean than your friend

world populations:

a. range is to subtract the smallest number from the largest

803 - 31 = 772

b. mean is adding it all up and dividing by how many numbers there are:

803 + 487 + 348 + 368 + 730 + 31 = 2767 / 6 = 461.17

median is arranging them in order and finding the middle #

31, 348, 368, 487, 730, 803

368 + 487 = 855 / 2 = 427.5

mean = 461.17

median = 427.5

mode = no mode

tomato plants:

a. range is subtracting the smallest from the largest

52 - 36 = 16

b. mean is adding all the #s and dividing by how many #s there are:

36 + 45 + 52 + 40 + 38 +41 + 50 + 48 = 350 / 8 = 43.75

median is the middle from least to greatest

36, 38, 40, 41, 45, 48, 50, 52

41 + 45 = 86 / 2 = 43

mean = 43.75

median = 43

mode = no mode

hope this helps! :)

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3 years ago

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3 years ago

What’s the answer to this

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2 years ago

The average production cost for major movies is 57 million dollars and the standard deviation is 22 million dollars. Assume the

Degger [83]

Using the normal distribution, we have that:

The distribution of X is $X \approx (57,22)$ .

The distribution of $\mathbf{\bar{X}}$ is $\bar{X} \approx (57, 5.3358)$ .

0.0597 = 5.97% probability that a single movie production cost is between 55 and 58 million dollars.

0.2233 = 22.33% probability that the average production cost of 17 movies is between 55 and 58 million dollars. Since the sample size is less than 30, assumption of normality is necessary.

<h3>Normal Probability Distribution</h3>

The z-score of a measure X of a normally distributed variable with mean $\mu$ and standard deviation $\sigma$ is given by:

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

The z-score measures how many standard deviations the measure is above or below the mean.

Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

In this problem, the parameters are given as follows:

$\mu = 57, \sigma = 22, n = 17, s = \frac{22}{\sqrt{17}} = 5.3358$

Hence:

The distribution of X is $X \approx (57,22)$ .

The distribution of $\mathbf{\bar{X}}$ is $\bar{X} \approx (57, 5.3358)$ .

The probabilities are the p-value of Z when X = 58 subtracted by the p-value of Z when X = 55, hence, for a single movie:

X = 58:

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

$Z = \frac{58 - 57}{22}$

Z = 0.05.

Z = 0.05 has a p-value of 0.5199.

X = 55:

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

$Z = \frac{55 - 57}{22}$

Z = -0.1.

Z = -0.1 has a p-value of 0.4602.

0.5199 - 0.4602 = 0.0597 = 5.97% probability that a single movie production cost is between 55 and 58 million dollars.

For the sample of 17 movies, we have that:

X = 58:

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

$Z = \frac{58 - 57}{5.3358}$

Z = 0.19.

Z = 0.19 has a p-value of 0.5753.

X = 55:

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

$Z = \frac{55 - 57}{5.3358}$

Z = -0.38.

Z = -0.38 has a p-value of 0.3520.

0.5753 - 0.3520 = 0.2233 = 22.33% probability that the average production cost of 17 movies is between 55 and 58 million dollars. Since the sample size is less than 30, assumption of normality is necessary.

More can be learned about the normal distribution at brainly.com/question/4079902

#SPJ1

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1 year ago

Round of 1,546 to the nearest thousands

EleoNora [17]

Answer:

1,546 ≈ 2,000

Hope this helps :)

1. To round to nearest thousands, first look at the hundreds place if the number is 1,2,3,4 then round down. If the number is 5,6,7,8,9 then round up.

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1 year ago