It is possible to obtain a ""quick-and-dirty"" estimate of the mean of a normal distribution from the 50th percentile value on a
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Using the z-distribution, as we are working with a proportion, it is found that 1016 constituents are required.
<h3>What is a confidence interval of proportions?</h3>A confidence interval of proportions is given by:
In which:
is the sample proportion.
- z is the critical value.
- n is the sample size.
The margin of error is given by:
In this problem, we have a 95% confidence level, hence, z is the value of Z that has a p-value of
, so the critical value is z = 1.96.
The estimate is of , while the margin of error is of M = 0.03, hence solving for n we find the minimum sample size.
Rounding up, 1016 constituents are required.
More can be learned about the z-distribution at brainly.com/question/25890103
Using Chebyshev's Theorem, considering a standard deviation of 40, we have that at least 75% of the students scored between 360 and 520.
<h3>What does Chebyshev’s Theorem state?</h3>When we have no information about the population distribution, Chebyshev's Theorem is used. It states that:
- At least 75% of the measures are within 2 standard deviations of the mean.
- At least 89% of the measures are within 3 standard deviations of the mean.
- An in general terms, the percentage of measures within k standard deviations of the mean is given by
.
In this problem, considering a standard deviation of 40, we have that:
440 - 2 x 40 = 360.
440 + 2 x 30 = 520.
Within 2 standard deviations of the mean, no information about the distribution, hence, at least 75% of the students scored between 360 and 520.
More can be learned about Chebyshev's Theorem at brainly.com/question/25303620