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Bas_tet [7]
3 years ago
14

How do i find quartiles, 1,2 and 3

Mathematics
2 answers:
AlekseyPX3 years ago
5 0

Answer:

see explanation

Step-by-step explanation:

To obtain the quartiles arrange the data in ascending order

38, 47, 54, 60, 67, 68, 76, 77, 84, 89

Q₂ is the median

The median is the middle value in the data set. If there is no exact middle value then it is the average of the 2 values either side of the middle

Q₂ = \frac{67+68}{2} = 67.5

Q₁ is the lower quartile and is the middle value of the data to the left of the median

Q₁ = 54

Q₃ is the upper quartile and is the middle value of the data to the right of the median

Q₃ = 77

Alexus [3.1K]3 years ago
3 0

Lower quartile=25% of total frequency/numbers in the data set

Median=50% of total frequency/numbers in the data set

Upper quartile=75% of total frequency/numbers in the data set

Hope this helps:)

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Parts being manufactured at a plant are supposed to weigh 65 grams. Suppose the distribution of weights has a Normal distributio
andrew-mc [135]

Answer:

0.64%.

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 normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score 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 p-value, 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.

Mean of 75 grams and a standard deviation of 22 grams.

This means that \mu = 75, \sigma = 22

Sample of 144:

This means that n = 144, s = \frac{22}{\sqrt{144}} = 1.8333

More than 80 or less than 70:

Both are the same distance from the mean, so we find one probability and multiply by 2.

The probability that it is less than 70 is the pvalue of Z when X = 70. So

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

By the Central Limit Theorem

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

Z = \frac{70 - 75}{1.8333}

Z = -2.73

Z = -2.73 has a pvalue of 0.0032

2*0.0032 = 0.0064.

0.0064*100% = 0.64%

The probability is 0.64%.

3 0
3 years ago
2,803 x 406. Estimate the product first. ≈ ________________ × ________________
Advocard [28]

Answer:

approximately 1,148,000

Step-by-step explanation:Round 2,803 to 2,800 and 406 to 410 then multiply 2,800 by 410

8 0
3 years ago
A store pays $32.00 for each video game. The store’s percent of mark up is 75%. Which shows the retail price of the video game?
Black_prince [1.1K]
The answer is $56 because 75% of $32 is 24 so you add 24 to 32.
5 0
3 years ago
HELP ME pleaseeeee! I'LL GIVE YOU BRANILEST!
earnstyle [38]

Answer:

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Step-by-step explanation:

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5 0
3 years ago
Read 2 more answers
The final exam of a particular class makes up 40% of the final grade, and Moe is failing the class with an average (arithmetic m
amm1812

Answer:

%82.5

Step-by-step explanation:

  1. The final exam of a particular class makes up 40% of the final grade
  2. Moe is failing the class with an average (arithmetic mean) of 45% just before taking the final exam.

From point 1 we know that Moe´s grade just before taking the final exam represents 60% of the final grade. Then, using the information in the point 2 we can compute Moe´s final grade as follows:

FG=0.40*FE+0.60*0.45,

where FG is Moe´s Final Grade and FE is Moe´s final exam grade. Then,

\frac{ FG-0.60*0.45}{0.40}=FE.

So, in order to receive the passing grade average of 60% for the class Moe needs to obtain in his exam:

FE=\frac{ 0.60-0.60*0.45}{0.40}=0.825

That is, he need al least %82.5 to obtain a passing grade.

6 0
3 years ago
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