I will give you BRAINLIEST for the correct answer

12.4. Measure of Dispersion 2076 Q.No. 15 Find the standard deviation of: 4, 6, 8, 10, 12.

Alenkinab [10]

Answer:

Standard deviation of given data = 3.16227

Step-by-step explanation:

Step(i):-

Given sample size 'n' = 5

Given data 4, 6,8,10,12

$Mean = \frac{4+6+8+10+12}{5} = 8$

Mean of the sample x⁻ = 8

Standard deviation of the sample

$S.D = \sqrt{\frac{Sum(x-x^{-} )^{2} }{n-1}}$

Step(ii):-

Given data

x : 4 6 8 10 12

x-x⁻ : 4 - 8 6-8 8-8 10-8 12-8

(x-x⁻) : -4 -2 0 2 4

(x-x⁻)² : 16 4 0 4 16

$S.D = \sqrt{\frac{Sum(x-x^{-} )^{2} }{n-1}}$

$S.D = \sqrt{\frac{16+4+0+4+16}{4}}$

S.D = √10 = 3.16227

Final answer:-

The standard deviation = 3.16227

8 0

2 years ago

Which shows the expression below in simplified form? (2.2 × 1012) + (1.7 × 109)

AleksAgata [21]

(2.2 × 1012) + (1.7 × 109) = 2411.7

7 0

3 years ago

Read 2 more answers

Find the value of x and y that will make each quadrilateral a parallelogram.

Tatiana [17]

Answer:

the opposite Angles sum up to give 180

Step-by-step explanation:

3y +3y = 180

y=30

if y = 30 the x = 55

are you clear?

7 0

3 years ago

Read 2 more answers

Verify that the segments are parallel

kozerog [31]

Answer:

We know that lines AB and CD are parallel because if you expand them, we know they will never touch.

Step-by-step explanation:

4 0

3 years ago

Average precipitation for the first 7 months of the year, the average precipitation in toledo, ohio, is 19.32 inches. if the ave

Colt1911 [192]

Part A:

The probability that a normally distributed data with a mean, μ and standard deviation, σ is greater than a given value, a is given by:

$P(x\ \textgreater \ a)=1-P(x\ \textless \ a)=1-P\left(z\ \textless \ \frac{a-\mu}{\sigma}\right)$

Given that the average precipitation in Toledo, Ohio for the past 7 months is 19.32 inches with a standard deviation of 2.44 inches, the probability that a randomly selected year will have precipitation greater than 18 inches for the first 7 months is given by:

$P(x\ \textgreater \ 18)=1-P(x\ \textless \ 18) \\ \\ =1-P\left(z\ \textless \ \frac{18-19.32}{2.44}\right) \\ \\ =1-P(z\ \textless \ -0.5410) \\ \\ =1-0.29426=\bold{0.7057}$

Part B:

The probability that an n randomly selected samples of a normally distributed data with a mean, μ and standard deviation, σ is greater than a given value, a is given by:

$P(x\ \textgreater \ a)=1-P(x\ \textless \ a)=1-P\left(z\ \textless \ \frac{a-\mu}{\frac{\sigma}{\sqrt{n}}}\right)$

Given that the average precipitation in Toledo, Ohio for the past 7 months is 19.32 inches with a standard deviation of 2.44 inches, the probability that 5 randomly selected years will have precipitation greater than 18 inches for the first 7 months is given by:

 $P(x\ \textgreater \ 18)=1-P(x\ \textless \ 18) \\ \\ =1-P\left(z\ \textless \ \frac{18-19.32}{\frac{2.44}{\sqrt{5}}}\right) \\ \\ =1-P(z\ \textless \ -1.210) \\ \\ =1-0.1132=\bold{0.8868}$

7 0

3 years ago