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3 years ago

What is the median of the data 21,22,23,24,25

Mathematics

2 answers:

Arlecino [84]3 years ago

5 0

✧･ﾟ: *✧･ﾟ:* *:･ﾟ✧*:･ﾟ✧

Hello!

✧･ﾟ: *✧･ﾟ:* *:･ﾟ✧*:･ﾟ✧

❖ The median of the data is 23.

You keep crossing out numbers until you get to the middle point. First, you cross of the beginning and the end which is 21 and 25. You then cross out 22 and 24. The remaining number is 23. When you are working with an even amount of numbers, you are left with 2 numbers so you add those 2 numbers and divide by 2.

~ ʜᴏᴘᴇ ᴛʜɪꜱ ʜᴇʟᴘꜱ! :) ♡

~ ᴄʟᴏᴜᴛᴀɴꜱᴡᴇʀꜱ

nydimaria [60]3 years ago

5 0

Answer is in the image provided.

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Suppose monthly rental prices for a one-bedroom apartment in a large city has a distribution that is skewed to the right with a

omeli [17]

Answer:

a) Nothing, beause the distribution of the monthly rental prices are not normal.

b) 1.43% probability that the sample mean rent price will be greater than $900

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 $\sigma$ , the sample means with size n of at least 30 can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$

(a) Suppose a one-bedroom rental listing in this large city is selected at random. What can be said about the probability that the listed rent price will be at least $930?

Nothing, beause the distribution of the monthly rental prices are not normal.

(b) Suppose a random sample 30 one-bedroom rental listing in this large city will be selected, the rent price will be recorded for each listing, and the sample mean rent price will be computed. What can be said about the probability that the sample mean rent price will be greater than $900?

Now we can apply the Central Limit Theorem.

$\mu = 880, \sigma = 50, n = 30, s = \frac{50}{\sqrt{30}} = 9.1287$

This probability is 1 subtracted by the pvalue of Z when X = 900.

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

By the Central Limit Theorem

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

$Z = \frac{900 - 880}{9.1287}$

$Z = 2.19$

$Z = 2.19$ has a pvalue of 0.9857

1 - 0.9857 = 0.0143

1.43% probability that the sample mean rent price will be greater than $900

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3 years ago

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melamori03 [73]

Answer:

The answer to your question is a) 4/12 b) 5/12

Step-by-step explanation:

Data

Lettuce = 1/12

Carrots = 3/12

Corn = 4/12

Red radishes = ?

The total of the garden is 12/12

Red radishes = 12/12 - 1/12 - 3/12 - 4/12

= 12/12 - 8/12

= 4/12

4/12 of the garden are red radishes.

Corn of lettuce = 1/12 + 4/12

= 5/12

5/12 of the garden are corn or lettuce.

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marysya [2.9K]

Answer:

Step-by-step explanation:

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