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2 years ago

It has been reported that the

Mathematics

1 answer:

Rainbow [258]2 years ago

7 0

If the standard deviation is 9 and sample size is 100 then the standard error is 0.9.

Given that the mean is 80,standard deviation be 9 and sample size be 100.

We are required to find the standard error of the given data.

The standard error of a statistic is basically the standard deviation of its sampling distribution or an estimate of that standard deviation.

Standard error=σ/ $\sqrt{n}$

in which σ is standard deviation and n is sample size.

Here standard deviation is 9 and sample size is 100.

Standard deviation=9/ $\sqrt{100}$

=9/10

=0.9

Hence if the standard deviation is 9 and sample size is 100 then the standard error is 0.9.

Learn more about standard error at brainly.com/question/1191244

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You might be interested in

What is the expanded notation for 2,909? Numbers may be used Once,More than once, or not at all.

defon

(2 x 1,000) + (9 x 100) + (9 x 1)
Always check the place values by looking at the number (2,909) and you can figure out what number to multiply it by. Guess and check if you can’t recognize it right away.

3 0

3 years ago

a company is expanding its building area from 17,500 square feet to 21,175 square feet. what is the percentage increase of the a

NikAS [45]

Let's make an equation.

19,000x=24,320

This equation shows the percentage increase, which is x. Once we have x, we need to subtract 1 from it since in the equation the 1 shows that it's increeasing. All we need to do is divide both sides by 19,000.

24,320÷19,000

1,28-1=0.28; we need to multiply by 100 to find the increase.

0.28×100=28

So, the percentage increase is 28%.

3 0

3 years ago

According to a Los Angeles Times study of more than 1 million medical dispatches from to , the response time for medical aid var

-BARSIC- [3]

Answer:

A)Mean :10.65

Median =10.7

Mode : 10.7

B)Range = 3.5

Standard deviation :0.89916

C)The response time of 8.3 minutes should be considered an outlier in comparison to the other response times

Step-by-step explanation:

Data : 11.8 ,10.3, 10.7, 10.6, 11.5, 8.3, 10.5, 10.9, 10.7, 11.2

$Mean = \frac{\text{Sum of all observations}}{\text{No. of observations}}\\Mean = \frac{11.8 +10.3+ 10.7+ 10.6+ 11.5+ 8.3+ 10.5+ 10.9+ 10.7+ 11.2}{10}$

Mean = 10.65

Median: The mid value of the data

Data in ascending order

8.3

10.3

10.5

10.6

10.7

10.9

11.2

11.5

11.8

n=10(even)

$Median = \frac{(\frac{n}{2})\text{th term}+(\frac{n}{2}+1)\text{th term}}{2}\\Median = \frac{(\frac{10}{2})\text{th term}+(\frac{10}{2}+1)\text{th term}}{2}\\Median = \frac{10.7+10.7}{2}\\Median = 10.7$

Mode : the most occurring frequency

10.7 is occurring twice while others are occurring once

So, Mode is 10.7

B) Range = Maximum - Minimum=11.8-8.3=3.5

Standard deviation : $\sqrt{\frac{\sum(x-\bar{x})^2}{n}}$

Standard deviation : $\sqrt{\frac{(11.8-10.65)^2+(10.3-10.65)^2+......+(10.9-10.65)^2+(10.7-10.65)^2+(11.2-10.65)^2}{10}}$

Standard deviation :0.89916

8.3

,10.3

,10.5

,10.6

,10.7

,10.9

,11.2

,11.5

,11.8

For Q1 ( Median of lower quartile )

8.3

,10.3

,10.5

,10.6

,10.7

$Median = \frac{n+1}{2}\text{th term} =\frac{5+1}{2}=3 \text{rd term}=10.5$

For Q3( Median of Upper quartile )

10.7

,10.9

,11.2

,11.5

,11.8

$Median = \frac{n+1}{2}\text{th term} =\frac{5+1}{2}=3 \text{rd term}=11.2$

IQR = Q3-Q1=11.2-10.5=0.7

Range :(Q1-1.5IQR, Q3-1.5IQR)

Range : $(10.5-1.5 \times 0.7, 11.2-1.5 \times 0.7)$

Range :(9.45, 10.15)

8.3 does not lie in this interval

So, the response time of 8.3 minutes should be considered an outlier in comparison to the other response times

3 0

3 years ago

F and H are sets of real numbers defined as follows.

Marina CMI [18]

Answer:

$(a)F \cup H=(3, \infty)\\(b)F \cap H=(6, \infty)$

Step-by-step explanation:

Given the sets F and H of real numbers defined as given:

F={w | w>3}

H={w | w>6}

To write a set in interval notation, we use these brackets, ( ) for < or > while we use [ or ] for $\leq $ or \geq$ .

Since our sets are defined by strict inequalities(< or >)

In interval notation

$F=(3,\infty), H=(6,\infty)$

$(a) F \cup H =(3,\infty) \cup (6,\infty) \\\\F \cup H=(3, \infty)\\\\(b) F \cap H =(3,\infty) \cap (6,\infty) \\\\F \cap H=(6, \infty)$

8 0

3 years ago

A rectangle and regular hexagon are shown below. Find the value of x that makes the polygons above have the same perimeter

pav-90 [236]

The answer is x=3
because 3x2 is 6 and 6x6 is 36 because a hexagon has 6 sides, and 3x3 is 9+3 is 12 and 12+12 is 24 and 3+3 is 6 so 6+6 is 12 and is you add that to 24 then that’s 36 aswell.

6 0

4 years ago