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3 years ago

A. For exercises 1 and 2, a set of data has a normal distribution with a mean of 120 and a standard deviation of 10.

1 answer:

4 0

A1.) the interval about the mean within which 90% of the data lie = 120 + or - 1.645(10) = 120 + or - 16.45 = 120 - 16.45 to 120 + 16.45 = 103.55 to 136.45

2.) P(100 < X < 140) = P(X < 140) - P(X < 100) = P(z < (140 - 120)/10) - P(z < (100 - 120)/10) = P(z < 2) - P(z < -2) = P(z < 2) - [1 - P(z < 2)] = 2P(z < 2) - 1 = 2(0.97725) - 1 = 1.9545 - 1 = 0.9545 = 95.5%

B1.) the interval about the sample mean that has a 1% level of confidence is 500 + or - 2.575(80/√1000) = 500 + or - 7 = 500 - 7 to 500 + 7 = 493 to 507

2.) 2P(z < a) - 1 = 0.90
2P(z < a) = 1.90
P(z < a) = 0.95
a = 1.645
(b - 500)/(80/√1000) = 1.645
b - 500 = 4
b = 500 + 4 = 504
The interval about the sample mean such that the probability is 0.90 that the mean number lies within the interval is 496 - 504.

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Answer:

The length of side x is $x = \sqrt{6}$

Step-by-step explanation:

Pythagorean Theorem:

In a right triangle, with sides a and b, and hypothenuse h, we have that:

$a^2 + b^2 = h^2$

In this question:

The vertical line on the sides means that both are equal, that is, measuring x, so $a = b = x$ .

Hypothenuse is the square root of 12. So $h = \sqrt{12}$

$x^2 + x^2 = (\sqrt{12})^2$

$2x^2 = 12$

$x^2 = \frac{12}{2}$

$x^2 = 6$

$x = \sqrt{6}$

The length of side x is $x = \sqrt{6}$

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2 years ago

Coach Evans recorded the height, in inches of each player on his team. The results are shown.

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Answer:

Step-by-step explanation:

Given:

Team heights (inches):

61, 57, 63, 62, 60, 64, 60, 62, 63

To find: IQRs (interquartile ranges) of the heights for the team

Solution:

A quartile divides the number of terms in the data into four more or less equal parts that is quarters.

For a set of data, a number for which 25% of the data is less than that number is known as the first quartile $(Q_1)$

For a set of data, a number for which 75% of the data is less than that number is known as the third quartile $(Q_3)$

Terms in arranged in ascending order:

$57,60,60,61,62,62,63,63,64$

Number of terms = 9

As number of terms is odd, exclude the middle term that is 62.

$Q_1$ is median of terms $57,60,60,61$

Number of terms (n) = 4

Median = $\frac{(\frac{n}{2})^{th} +(\frac{n}{2}+1)^{th} }{2} =\frac{2^{nd}+3^{rd}}{2} =\frac{60+60}{2}=\frac{120}{2}=60$

So, $Q_1=60$

So, 25% of the heights of a team is less than 60 inches

$Q_3$ is the median of terms $62,63,63,64$

Median = $\frac{(\frac{n}{2})^{th} +(\frac{n}{2}+1)^{th} }{2} =\frac{2^{nd}+3^{rd}}{2} =\frac{63+63}{2}=\frac{126}{2}=63$

So, $Q_3=63$

So, 75% of the heights of a team is less than 63 inches

Interquartile range = $Q_3-Q_1=63-60=3$

The interquartile range is a measure of variability on dividing a data set into quartiles.

The interquartile range is the range of the middle 50% of the terms in the data.

So, 3 is the range of the middle 50% of the heights of the students.

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2 years ago

Which of the following describes the image of a figure after a dilation that has a scale factor between zero and one?

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Answer:

When a figure or shape is dilated , the size of the figure either enlarges or shrank without changing its actual shape.

Shape or figure after dilation is similar to original figure.

If the scale or factor of dilation is greater than 1 the shape enlarges.

If the scale factor lies between 0<scale factor<1, then the figure shrinks i.e size reduces.

So out of all the options given →It has the same shape as the original and is smaller than the original figure.

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3 years ago

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Answer:

and why did teacher ate it

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Answer:

We need a sample size of at least 1161.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.