A population has a mean of 300 and a standard deviation of 70. Suppose a sample of size 125 is selected and is used to estimate . Use z-table. What is the probability that the sample mean will be within +/- 8 of the population mean (to 4 decimals)? (Round z value in intermediate calculations to 2 decimal places.) What is the probability that the sample mean will be within +/- 17 of the population mean (to 4 decimals)? (Round z value in intermediate calculations to 2 decimal places.)

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MrSmoot MrSmoot · Answer 1 · 2019-04-15T21:37:01+02:00

Answer:

a: 0.8884

b: 0.9934

Step-by-step explanation:

We have µ = 300 and σ = 70. The sample size, n = 125.

For the sample to be within 8 units of the population mean, we would have sample values of 292 and 308, so we want to find:

P(292 < x < 308).

We need to find the z-scores that correspond to these values using the given data. See attached photo 1 for the calculation of these scores.

We have P(292 < x < 308) = 0.8884

Next we want the probability of the sample mean to be within 17 units of the population mean, so we want the values from 283 to 317. We want to find

P(283 < x < 317)

We need to find the z-scores that correspond to these values. See photo 2 for the calculation of these scores.

We have P(283 < x < 317) = 0.9934

codiepienagoya codiepienagoya · Answer 2 · 2022-01-25T14:41:00+01:00

Following are the calculation to the given points:

Using central limit theorem:

[tex]\to \mu_{\bar{x}}= \mu = 300\\\\ \to \sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}=\frac{70}{\sqrt{125}}= 6.261\\\\[/tex]

For point a)

Within [tex]\pm 8[/tex] population mean:[tex]\to P(\mu-8<\bar{x}<\mu+8)=P\left ( \frac{\mu-8-\mu}{6.261}Within [tex]\pm 17[/tex]

For point b)

Within [tex]\pm 17[/tex] population mean:

[tex]\to P(\mu-17<\bar{x}<\mu+17)=P\left ( \frac{\mu-17-\mu}{6.261}

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