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2 years ago

Which one of these statistics is unaffected by outliers?Mean Interquartile range Standard deviation Range

Mathematics

1 answer:

kirill115 [55]2 years ago

5 0

Answer:

Since Interquartile Range consider the middle values in the data set, i.e. 50%, it is not affected by outliers or extreme values. While other statistics mean, standard deviation and range are all affected by outliers or extreme values.

Step-by-step explanation:

hope this helps you :) can you help me please northc04 :(

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luis has a cooler filled with cans of soda. For every 5 cans of coke there are 3 cans of sprite. If there are 25 cans of coke, h

faltersainse [42]

for every 5 is 3, so 25*3= equals 75,

5 0

2 years ago

Read 2 more answers

Use the position function s(t) = −16t² + 400, which gives the height (in feet) of an object that has fallen for t seconds from a

skad [1K]

Answer:

160m/s

Step-by-step explanation:

The object can hit the ground when t = a; meaning that s(a) = s(t) = 0

So, 0 = -16a² + 400

16a² = 400

a² = 25

a = √25

a = 5 (positive 5 only because that's the only physical solution)

The instantaneous velocity is

v(a) = lim(t->a) [s(t) - s(a)]/[t-a)

Where s(t) = -16t² + 400

and s(a) = -16a² + 400

v(a) = Lim(t->a) [-16t² + 400 + 16a² - 400]/(t-a)

v(a) = Lim(t->a) (-16t² + 16a²)/(t-a)

v(a) = lim (t->a) -16(t² - a²)(t-a)

v(a) = -16lim t->a (t²-a²)(t-a)

v(a) = -16lim t->a (t-a)(t+a)/(t-a)

v(a) = -16lim t->a (t+a)

But a = t

So, we have

v(a) = -16lim t->a 2a

v(a) = -32lim t->a (a)

v(a) = -32 * 5

v(a) = -160

Velocity = 160m/s

7 0

2 years ago

It is claimed that 15% of the ducks in a particular region have patent schistosome infection. Suppose that seven ducks are selec

Marizza181 [45]

Answer:

(a) X follows a Binomial distribution

(b) (i) P(X ≥ 2) = 0.28348

P(X = 1) = 0.39601

P(X ≤ 3) = 0.98800

Step-by-step explanation:

(a) In this situation, the variable X equal to the number of ducks that are infected follows a <em>Binomial distribution</em> because we have:

n identical and independent events: The 7 ducks that are selected at random
2 possible results for every event: success and fail. We can call success if the duck is infected and fail if the duck is not infected.
A probability p of success and 1-p of fail: There is a probability p equal to 15% that the ducks have the infection and a probability of (100%-15%) that they don't.

(b) So, the probability that X ducks are infected is calculated as:

$P(x)=\frac{n!}{x!(n-x)!}*p^{x}*(1-p)^{n-x}$

$P(x)=\frac{7!}{x!(7-x)!}*0.15^{x}*(1-0.15)^{7-x}$

Then, Probability P(X = 1) is equal to:

$P(1)=\frac{7!}{1!(7-1)!}*0.15^{1}*(1-0.15)^{7-1}=0.3960$

At the same way, probability P(X ≥ 2) is equal to:

P(X ≥ 2) = P(2) + P(3) + P(4) + P(5) + P(6) + P(7)

P(X ≥ 2) = 0.2097 + 0.0617 + 0.0109 + 0.0011 + 0.00006 + 0.00002

P(X ≥ 2) = 0.28348

And probability P(X ≤ 3) is equal to:

P(X ≤ 3) = P(0) + P(1) + P(2) + P(3)

P(X ≤ 3) = 0.3206 + 0.3960 + 0.2097 + 0.0617

P(X ≤ 3) = 0.988

6 0

2 years ago

Read 2 more answers

4. Jason and Allison are hiking in the woods when they spot a rare owl in a tree. Jason stops and measures an angle of elevation

fenix001 [56]

Here is the correct format for the question.

Jason and Allison are hiking in the woods when they spot a rare owl in a tree. Jason stops and measures an angle of elevation of 22° 8’ 6”. At the same time, Allison is standing 48 feet closer to the tree, measuring an angle of elevation of 30° 40’ 30” to the owl. If Jason and Allison are the same height, their eyes 5 feet from the ground, find the height of owl in the tree.

Answer:

The height of the owl from the ground is: 67.1337 ft

Step-by-step explanation:

The first step we are meant to start with is by convert our angle of elevation into degree;

So; we have the given data

22° 8’ 6”

30° 40’ 30”

To degrees ; we get :

$22^o\, 8' \,6" = 22 +\frac{8}{60} +\frac{6}{3600}=22.135^o\\ \\ 30^o\, 40' \,30" = 30 +\frac{40}{60} +\frac{30}{3600}=30.675^o$

From the attached file below; we can see a diagrammatic representation of the two right angle triangles together showing the data set which include the position of the hikers , the owl on the tree and the angle of elevations.

The trigonometric equations derived from the right angled triangle can be illustrated as:

$tan(22.135^o)=\frac{T}{48+x} \\ \\tan(30.675^o)=\frac{T}{x}$

From above equation ;

solving for the value of T , then equating both equations to determine the value of x to find the the height of the owl in the tree; we have:

$T=(48+x)\,tan(22.135^o)\\ \\T=x\,\,tan(30.675^o)\\ \\(48+x)\,tan(22.135^o)=x\,\,tan(30.675^o)\\ \\ 48\,tan(22.135^o)+x\,\,tan(22.135^o)=x\,\,tan(30.675^o)\\ \\ 48\,tan(22.135^o)=x\,\,(tan(30.675^o)-tan(22.135^o))\\ \\ x=\frac{48\,tan(22.135^o)}{tan(30.675^o)-tan(22.135^o)} \\ \\x=104.75\,\,ft$

However; to find T; we have:

$T-x\,\,tan(30.675^o)\\ \\ T=(104.75\,ft)\,\,tan(30.675^o)\\ \\ T=62.1337\,\,ft$

From the question; if we take an integral look and an in depth understanding of the question; we will realize that the value of T = 62.1337 ft is the height the owl is relative to the eye level of the hikers; so if we want to determine the height of the owl from the ground, there is need to add the given (5 feet from the ground) to this number:

SO;

The height of the owl from the ground is: (62.1337 + 5) ft

= 67.1337 ft

5 0

2 years ago

An object was carefully weighed five times on a multiple-beam balance. The balance was zeroed each time. The masses were: 11.36

Eddi Din [679]

Answer:

11.38 g

0.014142

Step-by-step explanation:

part a

Mean = Sum of all results / No. of results

Mean = (11.36 + 11.37 + 11.40 + 11.38 + 11.39) / 5

Mean = 11.38 g = E (X)

part b

Var (X) = E (X^2) - (E(X))^2

S.d = sqrt (Var (X))

Compute E (X^2):

E (X^2) = (11.36^2 + 11.37^2 + 11.40^2 + 11.38^2 + 11.39^2) / 5

E (X^2) = 129.5046

(E(X))^2 = 11.38^2 = 129.5044

Var (X) = 129.5046 - 129.5044 = 0.0002

s.d = sqrt (0.0002) = 0.014142

3 0

2 years ago