Hey there Nikki G.,
Thank you for posting these awesome and thought-provoking probabilities and statistics questions.
With normal distributions, we make some assumptions in order to do statistics with numbers representing data.
Normal distribution means that the mean lies in the center overlapping the mode and median with the most data points in the center and the rest evenly spread out around it uniformly and symmetrically. 68% of datapoints like +/- 1 standard deviation from the mean, 95% lie within 2, and 99.7% lie within 3 standard deviations either above or below the mean. The calculation is interesting.
To estimate, you can think of the normal distribution as containing half of data points below the mean and half above it, or 0.5.
given that 2 points away from the mean is 102 if abvoe or 98 if below and it is normally distributed with a standard deviation of 15, one standard deviation below the mean is 85 and one above is 115, then you can estimate that about .49 data points like below 98 and about .49 lie above 102, so 1 - .49x2 = 1-.98 = approximately 0.02 probability that the sample mean of IQ will not differ from the population mean by more than 2 points.
(b) also assumes normal population
z = x - u / o
Probability of x within 0.5 inch of population mean of 15inch is asking P( 14.5in < x < 15.5in) = ?
with a standard deviation of 12 inches
we can convert 14.5 and 15.5 to z-scores:
14.5 - 15 / 12 = -0.5/12 = -1/24 = -0.042
15.5 - 15/12 = .5/12 = 1/24 = 0.042.
Find the probability within these z-scores.
