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Arlecino [84]
3 years ago
6

HELP...................................

Mathematics
1 answer:
satela [25.4K]3 years ago
4 0

Answer:

16

Step-by-step explanation:

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Construc t a 95% confidence interval for the population standard deviation σ of a random sample of 15 men who have a mean weight
GrogVix [38]

The 95% confidence interval for the population standard deviation will be,

$\sqrt{\frac{(n-1) s^{2}}{\chi^{2}} \frac{\alpha}{2}} & < \sigma < \sqrt{\frac{(n-1) s^{2}}{\chi^{2}}} \\

simplifying the equation, we get

$\sqrt{\frac{2551.5}{26.1}} & < \sigma < \sqrt{\frac{2551.5}{5.63}} \\9.887 & < \sigma < 21.288

The 95% confidence interval exists (9.887, 21.288).

<h3>What is the  95% of confidence interval?</h3>

A random sample of 15 men exists selected. The mean weight exists at $165.2 pounds and the standard deviation exists at $13.5 pounds. The population exists normally distributed.

So, $n=15, \bar{X}=165.2, s=13.5$

where n exists the sample size, $\bar{X}$ exists the sample size

s exists the sample standard deviation.

The degrees of freedom will be n - 1 i.e.15 - 1 = 14.

For a 95% confidence interval, the level of significance will be $\alpha=0.05$.

The 95% confidence interval for the population standard deviation will be,

$\sqrt{\frac{(n-1) s^{2}}{\chi^{2}} \frac{\alpha}{2}} & < \sigma < \sqrt{\frac{(n-1) s^{2}}{\chi^{2}}} \\

substitute the values in the above equation, we get

$\sqrt{\frac{(15-1)(13.5)^{2}}{\chi^{2}} 0.05} & < \sigma < \sqrt{\frac{(15-1)(13.5)^{2}}{\chi_{0}^{2}}} \\

$\sqrt{\frac{(15-1)(13.5)^{2}}{\chi_{0.975}^{2}}} & < \sigma < \sqrt{\frac{(15-1)(13.5)^{2}}{\chi_{0.025}^{2}}} \\

simplifying the above equation, we get

$\sqrt{\frac{2551.5}{26.1}} & < \sigma < \sqrt{\frac{2551.5}{5.63}} \\9.887 & < \sigma < 21.288

The 95% confidence interval exists (9.887, 21.288).

To learn more about confidence interval refer to:

brainly.com/question/14825274

#SPJ4

6 0
2 years ago
What is the answer to this question
Gre4nikov [31]
It’s 4.08x10 to the 6 power. Which is answer A
6 0
3 years ago
Solve for L:<br> P=2L+2W
Korvikt [17]

Answer:

L=\frac{1}{2}P-W

Step-by-step explanation:

Let's solve for L.

P=2L+2W

Step 1: Flip the equation.

2L+2W=P

Step 2: Add -2W to both sides.

2L+2W+-2W=P+-2W

2L=P-2W

Step 3: Divide both sides by 2.

\frac{2L}{2}=\frac{P-2W}{2}

L=\frac{1}{2}P-W

Answer:

L=\frac{1}{2}P-W

4 0
3 years ago
it is known that a particular surgery has a 61% chance of being successful and caring a patient surgery is performed on five unr
Ganezh [65]

Answer:

about 3 of 5

Step-by-step explanation:

8 0
3 years ago
Can someone pls help me
WARRIOR [948]

Answer:

Total amount of fuel =200,000 litres

Amount of fuel used = 130,000 litres

Percentage of the fuel capacity used:

\frac{130000}{200000}  \times 100 =  \frac{13}{20}  \times 100 = 65\% \\

<em><u>65%</u></em> of the fuel is used on this flight.

5 0
3 years ago
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