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2 years ago

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A simple random sample of 50 items from a population with 7 resulted in a sample mean of 35.

1 answer:

Vilka [71]2 years ago

8 0

The 90% , 99% confidence interval for the population mean is 32.145 < $\rm \mu$ < 35.855 and 31.093 < $\rm \mu$ < 36.907

<h3>What is Probability ?</h3>

Probability is the study of likeliness of an event to happen.

It is given that

Total Population = 50

Mean = 35

The confidence interval is given by

$\rm CI = \mu + z \dfrac{s}{\sqrt n}$

$\rm \mu$ is the mean

z is the confidence level value

s is the standard deviation

n is the population width

(a) The 90% confidence interval for the population mean

90% $\rm \alpha / 2$ = 0.05

Z = 1.64

34 $\rm \pm$ 1.64 * 8 / √50

34 $\rm \pm$ 1.855

32.145 < $\rm \mu$ < 35.855

(b) The 99% confidence interval for the population mean

99% $\rm \alpha / 2$ = 0.005

Z=2.57

34 $\rm \pm$ 2.57 * 8 / √50

34 $\rm \pm$ 2.907

31.093 < $\rm \mu$ < 36.907

Therefore the confidence interval for population mean has been determined.

The complete question is

A simple random sample of 50 items from a population width =7 resulted in a sample mean of 35. If required, round your answers to two decimal places.

a. Provide a 90% confidence interval for the population mean

b. Provide a 99% confidence interval for the population mean

To know more about Probability

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Arrange the cones in order from lease volume to greatest volume

bearhunter [10]

Answer:

Volume of the cone in ascending order.

$V_{2}=270\pi\ units^{3}$

cone with DIAMETER of 18 & height of 10

cone with RADIUS of 10 & height of 9

cone with RADIUS of 11 & height of 9

cone with DIAMETER of 20 & height of 12

Step-by-step explanation:

Let $V_{2}. V_{3}. and\ V_{4}.$ be the volume of the cone.

Let d, r and h be the diameter, radius and height of the cone.

Given:

$d_{1} = 20\ and\ h_{1}=12$

$d_{2} = 18\ and\ h_{2}=10$

$r_{3} = 10\ and\ h_{3}=9$

$r_{4} = 11\ and\ h_{14}=9$

Arrange the cones in order from lease volume to greatest volume.

Solution:

The volume of the cone is given below.

$V=\pi r^{2} \frac{h}{3}$ ----------------(1)

where: r is radius of the base of cone.

and h is height of the cone.

The volume of the cone for $d_{1} = 20\ and\ h_{1}=12$

$r_{1} = \frac{d_{1}}{2}$

$r_{1} = \frac{20}{2}=10\ units$

$V_{1}=\pi (r_{1})^{2} \frac{h_{1}}{3}$

$V_{1}=\pi (10)^{2} \frac{12}{3}$

$V_{1}=\pi\times 100\times 4$

$V_{1}=400\pi\ units^{3}$

Similarly, for volume of the cone for $d_{2} = 18\ and\ h_{2}=10$

$r_{2} = \frac{d_{2}}{2}$

$r_{2} = \frac{18}{2}=9\ units$

$V_{2}=\pi (r_{2})^{2} \frac{h_{2}}{3}$

$V_{2}=\pi (9)^{2} \frac{10}{3}$

$V_{2}=\pi\times 81\times \frac{10}{3}$

$V_{2}=\pi\times 27\times 10$

$V_{2}=270\pi\ units^{3}$

Similarly, for volume of the cone for $r_{3} = 10\ and\ h_{3}=9$

$V_{3}=\pi (r_{3})^{2} \frac{h_{3}}{3}$

$V_{3}=\pi (10)^{2} \frac{9}{3}$

$V_{3}=\pi\times 100\times 3$

$V_{3}=\pi\times 300$

$V_{3}=300\pi\ units^{3}$

Similarly, for volume of the cone for $r_{4} = 11\ and\ h_{4}=9$

$V_{4}=\pi (r_{4})^{2} \frac{h_{4}}{3}$

$V_{4}=\pi (11)^{2} \frac{9}{3}$

$V_{4}=\pi\times 121\times 3$

$V_{4}=\pi\times 363$

$V_{4}=363\pi\ units^{3}$

So, the volume of the cone in ascending order.

$V_{2}=270\pi\ units^{3}$

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Answer: See below.

Step-by-step explanation:

The formatting is likely incorrect for these two options:

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I'll assume they were meant to be:

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In this case they would be non-linear, since y depends on the value of $x^{2}$

Use a "^" sign to indicate raised to a power: x^2 means $x^{2}$ .

y=(x−1)2 means y = 2x - 2

y=(x−1)^2 means y = $(x-1)^{2}$

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