One lap around the lake is 710 mile.

A random sample is taken and the sample size is 25. the sample is normally distributed, the sample mean is 80, and the standard

torisob [31]

The 90% confidence interval for the population mean is 78.1905<x<81.8095

<h3>Confidence interval</h3>

The formula for calculating the confidence interval is expressed as:

CI = x ± z*s/√n

Given the following parameters

mean "x" = 80

z = 1.645

s = 5.5

n = 25

Substitute

CI = 80 ± 1.645*5.5/√25

CI = 80 ± 1.8095

CI = {78.1905, 81.8095}

Hence the 90% confidence interval for the population mean is 78.1905<x<81.8095

Learn more on confidence interval here: brainly.com/question/2141785

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2 years ago

Suppose the scores on an exam are normally distributed with a mean μ = 75 points, and standard deviation σ = 8 points.

Artyom0805 [142]

Answer:

$P(X>69)=P(\frac{X-\mu}{\sigma}>\frac{69-\mu}{\sigma})=P(Z>\frac{69-75}{8})=P(Z>-0.75)$

And we can find this probability on this way:

$P(Z>-0.75)=1-P(Z$

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution to the problm

Let X the random variable that represent the scores on an exam of a population, and for this case we know the distribution for X is given by:

$X \sim N(75,8)$