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3 years ago

8. How do outliers affect the mean of a data set?

Mathematics

1 answer:

Zepler [3.9K]3 years ago

4 0

Answer:

it depends on whether the outlier is a high outlier or a low outlier

Step-by-step explanation:

The mean of a data set is the average of the data set. So if I had the data set {5, 2, 10} the average would be (5+2+10)/3 = 5.667. Now if I added an outlier like 10,000 it would increase the mean significantly. It changes it significantly but I cannot say whether it increases or decreases it without knowing the value of the outlier, since if I had the outlier -10,000 that would decrease the mean, but I had the outlier 10,000 it would increase it. So it depends on whether the outlier is a high outlier or a low outlier

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6th grade math :) help me please !

erma4kov [3.2K]

Answer:

a1- 2 terms

a2- 6x, 9y

a3- 6,9

b1-2 terms

b2- 7x, 12x

b3- 7,12

c1- 3 terms

c2- 5x^2, 2y, 3

c3- 5,2,3

Step-by-step explanation:

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4 years ago

Oranges cost $0.35 per pound and bananas cost $1.41 per

Anna [14]

She paid $5.669. Rounded to the nearest hundredth, $5.67. Rounded to the nearest whole number, $6.

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3 years ago

A random sample of 500 registered voters in Phoenix is asked if they favor the use of oxygenated fuels year-round to reduce air

Stells [14]

Answer:

a) 0.0853

b) 0.0000

Step-by-step explanation:

Parameters given stated that;

H₀ : p = 0.6

H₁ : p = 0.6, this explains the acceptance region as;

p° ≤ $\frac{315}{500}$ =0.63 and the region region as p°>0.63 (where p° is known as the sample proportion)

a).

the probability of type I error if exactly 60% is calculated as :

∝ = P (Reject H₀ | H₀ is true)

= P (p°>0.63 | p=0.6)

where p° is represented as pI in the subsequent calculated steps below

= P $[\frac{p°-p}{\sqrt{\frac{p(1-p)}{n}}} >\frac{0.63-p}{\sqrt{\frac{p(1-p)}{n}}} |p=0.6]$

= P $[\frac{p°-0.6}{\sqrt{\frac{0.6(1-0.6)}{500}}} >\frac{0.63-0.6}{\sqrt{\frac{0.6(1-0.6)}{500}}} ]$

= P $[Z>\frac{0.63-0.6}{\sqrt{\frac{0.6(1-0.6)}{500} } } ]$

= P [Z > 1.37]

= 1 - P [Z ≤ 1.37]

= 1 - Ф (1.37)

= 1 - 0.914657 ( from Cumulative Standard Normal Distribution Table)

≅ 0.0853

The probability of Type II error β is stated as:

β = P (Accept H₀ | H₁ is true)

= P [p° ≤ 0.63 | p = 0.75]

where p° is represented as pI in the subsequent calculated steps below

= P $[\frac{p°-p} \sqrt{\frac{p(1-p)}{n} } }\leq \frac{0.63-p}{\sqrt{\frac{p(1-p)}{n} } } | p=0.75]$

= P $[\frac{p°-0.6} \sqrt{\frac{0.75(1-0.75)}{500} } }\leq \frac{0.63-0.75}{\sqrt{\frac{0.75(1-0.75)}{500} } } ]$

= P $[Z\leq\frac{0.63-0.75}{\sqrt{\frac{0.75(1-0.75)}{500} } } ]$

= P [Z ≤ -6.20]

= Ф (-6.20)

≅ 0.0000 (from Cumulative Standard Normal Distribution Table).

6 0

3 years ago

What is the answer to this equation? (1.03×10to the 9)-(4.7×10to the 7)

STALIN [3.7K]

Answer:

x10000000 • (103x990000000 - 470)

—————————————————————————————————

100

7 0

3 years ago

What is 9-3 Divided by 1/3 + 1?

Eddi Din [679]

Answer:

9/2

Step-by-step explanation:

9-3=6 and (1/3+1)=4/3

The answer is 6/(4/3)=9/2

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3 years ago