Assume a normal distribution of the form N(100, 100), answer the following quest

ID: 3068503 • Letter: A

Question

Assume a normal distribution of the form N(100, 100), answer the following questions:

What proportion of the distribution falls within 1 standard deviation of the mean (e.g., within -1 and 1 standard deviation)?

What is the probability of a single draw from that distribution has a value greater than 115?

What is the range that captures the middle 95% of the population distribution?

If I randomly sample 10 observations from this distribution and calculate a mean, what is the probability that this mean is greater than 106?

If we center the sampling distribution on 100, then what is the range that will capture 95% of the means calculated from a sample of 20 observations?

Since the distribution is normal, so according to the empirical rule 68.3% of the distribution falls within 1 standard deviation of the mean.

Data given to us is:

Mean, m = 100

Standard deviation, S = 100^0.5 = 10

So, at X = 115, we have:

z = (X-m)/S = (115-100)/10 = 1.5

Looking at the cumulative z-table, we have:

P(X > 115) = P(z > 1.5) = 0.067

For the middle 95%, we have:

critical z-value, zc = 1.96

So,

Upper limit of range = m + 1.96*S = 100 + 1.96*10 = 119.6

Lower limit of range = m - 1.96*S = 100 - 1.96*10 = 80.4

So the range covering middle 95% of the population is 80.4 to 119.6

In this case we make use of sampling distribution of sample means.

For this distribution, standard error, SE = S/(n^0.5)

Sample size, n = 10

So,

SE = 10/(10^0.5) = 3.16

So, at X' = 106, we have:

z = (X'-m)/SE = (106-100)/3.16 = 1.89

Looking at the cumulative z-table, we have:

P(X' > 106) = P(z > 1.89) = 0.029

In this case, standard eror, SE = 10/(20^0.5) = 2.236

So in this case,

Upper limit of range = m + 1.96*SE = 100 + 1.96*2.236 = 104.39

Lower limit of range = m - 1.96*SE = 100 - 1.96*2.236 = 95.61

So the range covering middle 95% of the population is 95.61 to 104.39

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