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Ymorist [56]
3 years ago
13

The median is often a better representative of the central value of a data set when the data set: source is bimodal. has a high

standard deviation. is highly skewed. has no outliers.
Mathematics
1 answer:
Lorico [155]3 years ago
8 0
The median is a better representative of the central value of a data set when
the data set is highly skewed.

Note that
The median and average value only coincide when the distribution is perfectly normal.
Many outliers will also skew the distribution, and the median becomes a better representative.
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What he the value of n?<br> 3n+16=7n
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Answer:

The answer is 4

Step-by-step explanation:

3n+16=7n

collect like terms

-4n=-16

divide by -4

n=4

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What's 18/40 in simplest form
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18/40 divided by 2/2= 9/20

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It appears that people who are mildly obese are less active than leaner people. One study looked at the average number of minute
hoa [83]

Answer:

Obese people

Lower = \mu - 2\sigma = 373- 2(67) = 239

Upper = \mu + 2\sigma = 373+ 2(67) = 507

Lean People

Lower = \mu - 2\sigma = 526- 2(107) = 312

Upper = \mu + 2\sigma = 526+ 2(107) = 740

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

The empirical rule, also referred to as the three-sigma rule or 68-95-99.7 rule, is a statistical rule which states that for a normal distribution, "almost all data falls within three standard deviations (denoted by σ) of the mean (denoted by µ). Broken down, the empirical rule shows that 68% falls within the first standard deviation (µ ± σ), 95% within the first two standard deviations (µ ± 2σ), and 99.7% within the first three standard deviations (µ ± 3σ)".

Solution to the problem

Obese people

Let X the random variable that represent the minutes of a population (obese people), and for this case we know the distribution for X is given by:

X \sim N(373,67)  

Where \mu=373 and \sigma=67

On this case we know that 95% of the data values are within two deviation from the mean using the 68-95-99.7 rule so then we can find the limits liek this:

Lower = \mu - 2\sigma = 373- 2(67) = 239

Upper = \mu + 2\sigma = 373+ 2(67) = 507

Lean People

Let X the random variable that represent the minutes of a population (lean people), and for this case we know the distribution for X is given by:

X \sim N(526,107)  

Where \mu=526 and \sigma=107

On this case we know that 95% of the data values are within two deviation from the mean using the 68-95-99.7 rule so then we can find the limits liek this:

Lower = \mu - 2\sigma = 526- 2(107) = 312

Upper = \mu + 2\sigma = 526+ 2(107) = 740

The interval for the lean people is significantly higher than the interval for the obese people.

5 0
3 years ago
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