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Korolek [52]
3 years ago
6

According to a recent​ survey, the population distribution of number of years of education for​ self-employed individuals in a c

ertain region has a mean of 15.1 and a standard deviation of 4.8.
a. Identify the random variable X whose distribution is described here.
b. Find the mean and the standard deviation of the sampling distribution of x overbar for a random sample of size 36. Interpret them.
c. Repeat​ (b) for nequals144. Describe the effect of increasing n.
Mathematics
1 answer:
adell [148]3 years ago
6 0

Answer:

a)  X that is a distribution of number of years of education for​ self-employed individuals.

b) Mean = 15.1, Standard Deviation = 0.8

c) Mean = 15.1, Standard Deviation = 0.4

Step-by-step explanation:

We are given the following in the question:

The population distribution of number of years of education for​ self-employed individuals in a certain region has a mean of 15.1 and a standard deviation of 4.8

a) The random variable is X that is a distribution of number of years of education for​ self-employed individuals.

b) According to central limit theorem, as the sample size increases the distribution of means approaches a normal distribution.

Thus, \bar{x} has a normal distribution with

\text{Mean} = 15.1\\\text{Srtandard Deviation} = \displaystyle\frac{\sigma}{\sqrt{36}} = \frac{4.8}{\sqrt{36}} = 0.8

c) n = 144

\bar{x} has a normal distribution with

\text{Mean} = 15.1\\\text{Srtandard Deviation} = \displaystyle\frac{\sigma}{\sqrt{n}} = \frac{4.8}{\sqrt{144}} = 0.4

By increasing n, the standard deviation for distribution of mean reduced  by one half. Therefore, we see that quadrupling the sample size will reduce the standard deviation by one half.

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If edge of cube is reduced by 25%.find the percentage decrease in its total surface area.<br>​
vlada-n [284]
%75 I guess unless this is like highschool math
8 0
3 years ago
On a coordinate plane, a square and a point are shown. The square has points R prime (negative 8, 1), S prime (negative 4, 1), T
VladimirAG [237]

Answer:

(–1, –6)

Step-by-step explanation:

Given that S(3, –5) was translated to S'(–4, 1), the transformation rule is (x-7, y+6). Then, the coordinates of square RSTU are:

R'(–8, 1) -> (-8+7, 1-6) -> R(-1, -5)

S'(–4, 1) -> (-4+7, 1-6) -> S(3, -5)

T'(–4, –3) -> (-4+7, -3-6) -> T(3, -9)

U'(–8, –3) -> (-8+7, -3-6) -> U(-1, -9)

The point (–1, –6) lies on the segment RU

7 0
3 years ago
Read 2 more answers
NEED HELP FAST:
Irina-Kira [14]

Answer:

B- 5 5/6

Step-by-step explanation:

35 divided by 6 is 5 meaning there are 5 whole numbers in the fraction you then figure out how many 6 the remaining amount is which in this case is 5 so the answer is 5 5/6 hope this helps!

6 0
3 years ago
Assume that females have pulse rates that are normally distributed with a mean of μ=73.0 beats per minute and a standard deviati
Gennadij [26K]

Answer:

a. the probability that her pulse rate is less than 76 beats per minute is 0.5948

b. If 25 adult females are randomly​ selected,  the probability that they have pulse rates with a mean less than 76 beats per minute is 0.8849

c.   D. Since the original population has a normal​ distribution, the distribution of sample means is a normal distribution for any sample size.

Step-by-step explanation:

Given that:

Mean μ =73.0

Standard deviation σ =12.5

a. If 1 adult female is randomly​ selected, find the probability that her pulse rate is less than 76 beats per minute.

Let X represent the random variable that is normally distributed with a mean of 73.0 beats per minute and a standard deviation of 12.5 beats per minute.

Then : X \sim N ( μ = 73.0 , σ = 12.5)

The probability that her pulse rate is less than 76 beats per minute can be computed as:

P(X < 76) = P(\dfrac{X-\mu}{\sigma}< \dfrac{X-\mu}{\sigma})

P(X < 76) = P(\dfrac{76-\mu}{\sigma}< \dfrac{76-73}{12.5})

P(X < 76) = P(Z< \dfrac{3}{12.5})

P(X < 76) = P(Z< 0.24)

From the standard normal distribution tables,

P(X < 76) = 0.5948

Therefore , the probability that her pulse rate is less than 76 beats per minute is 0.5948

b.  If 25 adult females are randomly​ selected, find the probability that they have pulse rates with a mean less than 76 beats per minute.

now; we have a sample size n = 25

The probability can now be calculated as follows:

P(\overline X < 76) = P(\dfrac{\overline X-\mu}{\dfrac{\sigma}{\sqrt{n}}}< \dfrac{ \overline X-\mu}{\dfrac{\sigma}{\sqrt{n}}})

P( \overline X < 76) = P(\dfrac{76-\mu}{\dfrac{\sigma}{\sqrt{n}}}< \dfrac{76-73}{\dfrac{12.5}{\sqrt{25}}})

P( \overline X < 76) = P(Z< \dfrac{3}{\dfrac{12.5}{5}})

P( \overline X < 76) = P(Z< 1.2)

From the standard normal distribution tables,

P(\overline X < 76) = 0.8849

c. Why can the normal distribution be used in part​ (b), even though the sample size does not exceed​ 30?

In order to determine the probability in part (b);  the  normal distribution is perfect to be used here even when the sample size does not exceed 30.

Therefore option D is correct.

Since the original population has a normal distribution, the distribution of sample means is a normal distribution for any sample size.

5 0
3 years ago
Need help, please!! Don't anwser if you don't know it!!!!!!
Ne4ueva [31]

Answer:

C

Step-by-step explanation:

On the horizontal axis, as the discount (%) increases by 1(%), on the vertical axis; the sales increase (%) increases by 3(%).

When the discount is 10%, the sales increase is 32%.

When the discount is 20%, the sales increase is 62%.

20 - 10 = 10

10 × 3 = 30

32 + 30 = 62 (correct)

5 0
3 years ago
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