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torisob [31]
3 years ago
5

हरिहर काका किस घटना से व्यथित होकर ठाकुरबाड़ी चले गए?​ class 10 ​

Mathematics
1 answer:
Bas_tet [7]3 years ago
5 0

Answer:

what is that

Step-by-step explanation:

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Confidence Interval Mistakes and Misunderstandings—Suppose that 500 randomly selected recent graduates of a university were as
kvv77 [185]

Answer:

The correct 95% confidence interval is (8.4, 8.8).

Step-by-step explanation:

The information provided is:

n=500\\\bar x=8.6\\\sigma=2.2

(a)

The (1 - <em>α</em>)% confidence interval for population mean (<em>μ</em>) is:

CI=\bar x\pm z_{\alpha/2}\times \frac{\sigma}{\sqrt{n}}

The 95% confidence interval for the average satisfaction score is computed as:

8.6 ± 1.96 (2.2)

This confidence interval is incorrect.

Because the critical value is multiplied directly by the standard deviation.

The correct interval is:

8.6\pm 1.96 (\frac{2.2}{\sqrt{500}})=8.6\pm 0.20=(8.4,\ 8.6)

(b)

The (1 - <em>α</em>)% confidence interval for the parameter implies that there is (1 - <em>α</em>)% confidence or certainty that the true parameter value is contained in the interval.

The 95% confidence interval for the mean rating, (8.4, 8.8) implies that the true there is a 95% confidence that the true parameter value is contained in this interval.

The mistake is that the student concluded that the sample mean is contained in between the interval. This is incorrect because the population is predicted to be contained in the interval.

(c)

The (1 - <em>α</em>)% confidence interval for population parameter implies that there is a (1 - <em>α</em>) probability that the true value of the parameter is included in the interval.

The 95% confidence interval for the mean rating, (8.4, 8.8) implies that the true mean satisfaction score is contained between 8.4 and 8.8 with probability 0.95 or 95%.

Thus, the students is not making any misinterpretation.

(d)

According to the Central Limit Theorem if we have a population with mean <em>μ</em> and standard deviation <em>σ</em> and we take appropriately huge random samples (<em>n</em> ≥ 30) from the population with replacement, then the distribution of the sample mean will be approximately normally distributed.

In this case the sample size is,

<em>n </em>= 500 > 30

Thus, a Normal distribution can be applied to approximate the distribution of the alumni ratings.

7 0
3 years ago
A person is selected at random. If there are seven days in a week, what is the probability that the person was not born on a Tue
levacccp [35]

Answer:

5 out of 7.

Step-by-step explanation:

There are seven days in one week. They are asking what is the probability of the person chosen not being born in a Tuesday or Wednesday. There are five other days, so there are five other options.

I hope I helped you!

7 0
2 years ago
Mel has to put the greatest number of
Dmitry_Shevchenko [17]

Mel should use the least  common multiple to solve the problem

<u>Solution:</u>

Given, Mel has to put the greatest number of bolts and nuts in each box so each box has the same number of bolts and the same number of nuts.  

We have to find that should Mel use the greatest common factor or the least common multiple to solve the problem?

He should use least common multiple.

Let us see an example, suppose 12 bolts and nuts are to be fit in 6 boxes.

Then, if we took H.C.F of 12 and 6, it is 6, which means 6 bolts and nuts in each box, but, after filling 2 boxes with 6 bolts and nuts, there will be nothing left, which is wrong as remaining boxes are empty.

So the remaining method to choose is L.C.M.

Hence, he should use L.C.M method.

6 0
3 years ago
Mr. Sanchez buys 5 shirts and 3 ties for $34. At
Nastasia [14]

Answer:

y < x + 7

Step-by-step explanation:

7 0
2 years ago
PLZ HELP ME I will give u 10 points PLZZ HELP ME
zhannawk [14.2K]

Answer:

1000

7/100

Step-by-step explanation:

10×10×10is 100

8 0
2 years ago
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