Question

Mathematics

1 answer:

victus00 [196]3 years ago

3 0

Answer:

Step-by-step explanation:

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A test is used to assess readiness for college. In a recent year, the mean test score was 20.3 and the standard deviation was 4

Vlada [557]

Answer:

$\mu = 20.3$

$\sigma = 4.9$

And we can find the limits in order to consider values as significantly low and high like this:

$Low\leq \mu -2 \sigma= 20.3- 2*4.9 = 10.5$

$High\geq \mu +2 \sigma= 20.3+ 2*4.9 = 30.1$

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

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".

Solution to the problem

For this case we can consider a value to be significantly low if we have that the z score is lower or equal to - 2 and we can consider a value to be significantly high if its z score is higher tor equal to 2.

For this case we have the mean and the deviation given:

$\mu = 20.3$

$\sigma = 4.9$

And we can find the limits in order to consider values as significantly low and high like this:

$Low \leq \mu -2 \sigma= 20.3- 2*4.9 = 10.5$