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3 years ago

How does the mean absolute deviation (MAD) of the data in set 1 compare to the mean absolute deviation of the data in set 2?

Mathematics

1 answer:

frutty [35]3 years ago

4 0

Answer:

Mean Absolute deviation for (set 1): 4

Mean Absolute deviation for (set 2): 9

Set 2 has a lower (MAD) mean absolute deviation that Set 1.

Step-by-step explanation:

(set 1)

Mean: 80 + 82 + 90 = 252/3 = 84

84 - 80 = 4

84 - 82 = 2

84 - 90 = 6

Mean Absolute Deviation: 2 + 4 + 6 = 12/3 = 4

(set 2)

Mean: 60 + 80 + 82 + 90 = 312/4 = 78

78 - 60 = 18

78 - 80 = 2

78 - 82 = 4

78 - 90 = 12

Mean Absolute Deviation: 2 + 4 + 12 + 18 = 36/4 = 9

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3 years ago

A manufacturing process is designed to produce bolts with a 0.75-inch diameter. Once each day, a random sample of 36 bolts is se

larisa [96]

Answer:

0.0026 = 0.26% probability that the manufacturing line will be shut down unnecessarily

Step-by-step explanation:

We need to understand the normal probability distribution and the central limit theorem to solve this question.

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.

In this problem, we have that:

$\mu = 0.75, \sigma = 0.04, n = 36, s = \frac{0.04}{\sqrt{36}} = 0.0067$

What is the probability that the manufacturing line will be shut down unnecessarily?

Less than 0.73 or more than 0.77.

Less than 0.73

pvalue of Z when X = 0.73

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

By the Central Limit Theorem

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

$Z = \frac{0.73 - 0.75}{0.0067}$

$Z = -3$

$Z = -3$ has a pvalue of 0.0013

More than 0.77

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

$Z = \frac{0.77 - 0.75}{0.0067}$

$Z = 3$

$Z = 3$ has a pvalue of 0.9987

1 - 0.9987 = 0.0013

2*0.0013 = 0.0026

0.0026 = 0.26% probability that the manufacturing line will be shut down unnecessarily

3 0

2 years ago

Chuck earned $72.50 mowing lawns. The gas for the mower cost $11.25 and he bought a new string for the trimmer for $7.75 Write a

san4es73 [151]

We are given that:
Chunk earned : $72.5
Chunk spent: $11.25 for gas and $7.75 for new string.

We need to get the profit. Therefore, we will subtract the amount spent by Chunk from the amount he earned.

Therefore:
Chunk's profit = Amount earned - Amount spent
Chunk's profit = 72.5 - (11.25+7.75)
Chunk's profit = 72.5 - 19
Chunk's profit = $53.5

6 0

3 years ago

Chips Ahoy wants to perform a study to determine the number of chocolate chips in their cookies. To that end, they collected a s

Dahasolnce [82]

The formula for the confidence interval is given by

Sample mean + z*[σ/√n], and
Sample mean - z*[σ/√n]

We have:
Sample mean = 23.95
n = 40
σ = 2.55
z* for 99% confidence = 2.58

Substitute these values into the formula, we have

23.95 + (2.58)(2.55÷√40) = 24.99
23.95 - (2.58)(2.55÷√40) = 22.91

So the lower interval is 22.91 and the highest interval is 24.99

5 0

3 years ago

Please help me grades are do today

11111nata11111 [884]

Answer:

Step-by-step explanation:

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