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3 years ago

WILL GIVE BRAINLIEST

Mathematics

2 answers:

ad-work [718]3 years ago

4 0

The correct answer would be 1in, if you want to check and see if im right use a 12 inch ruler and check

hope this helps out.

9966 [12]3 years ago

3 0

I am guessing 1/4 inch

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A set of normally distributed data has a mean of 3.2 and a standard deviation of 0.7. Find the probability of randomly selecting

jenyasd209 [6]

Answer:

$P(Z> 3.13) = 0.000874$

Step-by-step explanation:

A set of normally distributed data has a mean of 3.2 and a standard deviation of 0.7. Find the probability of randomly selecting 30 values and obtaining an average greater than 3.6.

We can denote the population mean with the symbol $\mu$

According to the information given, the data have a population mean:

$\mu = 3.2$ .

The standard deviation of the data is:

$\sigma = 0.7$ .

Then, from the data, a sample of size $n = 30$ is taken.

We want to obtain the probability that the sample mean is greater than 3.6

If we call

$\mu_m$ to the sample mean then, we seek to find:

$P(\mu_m> 3.6)$

To find this probability we find the Z statistic.

$Z = \frac{\mu_m-\mu}{\sigma_{\mu_m}}$

Where:

Where $\sigma_{\mu_m}$ is the standard deviation of the sample

$\sigma_{\mu_m} = \frac{\sigma}{\sqrt{n}}$

$\sigma_{\mu_m} =\frac{0.7}{\sqrt{30}}\\\\\sigma_{\mu_m} = 0.1278$

$P(\frac{\mu_m-\mu}{\sigma_{\mu_m}}> \frac{3.6-3.2}{0.1278})$

Then:

$Z = \frac{3.6-3.2}{0.1278}\\\\Z = 3.13$

The probability sought is: $P(Z> 3.13)$

When looking in the standard normal probability tables for right tail we obtain:

$P(Z> 3.13) = 0.000874$

8 0

3 years ago

I need with this. Someone help please.

seraphim [82]

Answer:

In the ATTACHMENT

Step-by-step explanation:

For the first one, you can plug in multiple numbers of any kind for $x$ and start to form you line.

For the second one, know that the $y$ line is always horizontal. Find -3 on the plot that would be on the $y-axis$ and put that horizontal line on it.