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3 years ago

Which central measure is most affected by the outlier of this data set: 8, 10, 10, 11, 16, 17, 19, 21, 41 ?

Mathematics

1 answer:

tino4ka555 [31]3 years ago

6 0

Answer:

<h2>The mean is the most affected.</h2>

Step-by-step explanation:

In statistics, central measures are the mean, the median and mode.

The mean is the average, it indicates the value that the majority of the data set is falling to.

The median is the central value, literally. The value placed in the middle of the data set is the median.

The mode is the value with the highest frequency. In other words, the value that repeats the most.

So, in the given data set, the outlier is 41, which is a drastically greater value than the other ones in the data set. However, this oulier doesn't affect the mode or the median, because you can replace it with 22, and you would have the same median and the same mode.

Therefore, the central measure most affected is the mean, because this statistic measure uses all values in the data set, so if we change the outlier, the mean would change too.

So, the right answer is the mean.

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3 years ago

The operator of a pumping station has observed that demand for water during early afternoon hours has an approximately exponenti

melomori [17]

Answer:

a) 0.1496 = 14.96% probability that the demand will exceed 190 cfs during the early afternoon on a randomly selected day.

b) Capacity of 252.6 cubic feet per second

Step-by-step explanation:

Exponential distribution:

The exponential probability distribution, with mean m, is described by the following equation:

$f(x) = \mu e^{-\mu x}$

In which $\mu = \frac{1}{m}$ is the decay parameter.

The probability that x is lower or equal to a is given by:

$P(X \leq x) = \int\limits^a_0 {f(x)} \, dx$

Which has the following solution:

$P(X \leq x) = 1 - e^{-\mu x}$

The probability of finding a value higher than x is:

$P(X > x) = 1 - P(X \leq x) = 1 - (1 - e^{-\mu x}) = e^{-\mu x}$

The operator of a pumping station has observed that demand for water during early afternoon hours has an approximately exponential distribution with mean 100 cfs (cubic feet per second).

This means that $m = 100, \mu = \frac{1}{100} = 0.01$

(a) Find the probability that the demand will exceed 190 cfs during the early afternoon on a randomly selected day. (Round your answer to four decimal places.)

We have that:

$P(X > x) = e^{-\mu x}$

This is P(X > 190). So

$P(X > 190) = e^{-0.01*190} = 0.1496$

0.1496 = 14.96% probability that the demand will exceed 190 cfs during the early afternoon on a randomly selected day.

(b) What water-pumping capacity, in cubic feet per second, should the station maintain during early afternoons so that the probability that demand will exceed capacity on a randomly selected day is only 0.08?

This is x for which:

$P(X > x) = 0.08$