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3 years ago

What conclusions can you draw by comparing the mean and MAD repairs per vehicle for each type of Vehicle

1 answer:

5 0

The mean would tell you the average number of repairs per vehicle type. It is the total amount of repairs divided by the total number of vehicles.

The mean absolute deviation (MAD) describes how close the values are typically to the mean. It is an average of the distances the individual points are from the mean. Basically, it tells how spread out the data is.

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Need help with this problem please

MrMuchimi

I think she bought 23 feet

8 0

3 years ago

In a large company, the proportion of employees who were promoted during the last year was 0.10. If 100

Natasha2012 [34]

Answer:

0.0475

Step-by-step explanation:

We use the binomial approximation to the normal to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed distributions can be 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.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .

In a large company, the proportion of employees who were promoted during the last year was 0.10.

This means that $p = 0.1$

100 employees

This means that $n = 100$

Mean and standard deviation:

$\mu = E(X) = np = 100*0.1 = 10$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{100*0.1*0.9} = 3$

What is the probability that at least 15 of them were promoted during the last year?

This is $P(X \geq 15)$ , which is 1 subtracted by the pvalue of Z when X = 15. So

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

$Z = \frac{15 - 10}{3}$

$Z = 1.67$

$Z = 1.67$ has a pvalue of 0.9525

1 - 0.9525 = 0.0475.

0.0475 is the answer.

6 0

3 years ago

Help me on this question

olganol [36]

Answer:

see below

Step-by-step explanation:

All of the given data sets have x-values that are sequential with a difference of 1. That makes it easy to determine the sort of sequence the y-values make.

first choice: the y-values have a common difference of -2. This will be matched by a linear model.

second choice: the y-values have a common difference of +2. Again, this will be matched by a linear model.

third choice: the y-values have a common ratio of -2. This will be matched by an exponential model.

fourth choice: the y-value differences are 3, 5, 7, increasing by a constant amount (2). This is characteristic of a sequence that has a quadratic model.