The mean and standard deviation. If the data really are normally distributed, all other statistics are redundant.
A particular fruit's weights are normally distributed, with a mean of 760 grams and a standard deviation of 15 grams. If you pick one fruit at random, what is the probability that it will weigh between 722 grams and 746 grams-----A particular fruit's weights are normally distributed, with a mean of 567 grams and a standard deviation of 25 grams.
Anything that is normally distributed has certain properties. One is that the bulk of scores will be near the mean and the farther from the mean you are, the less common the score. Specifically, about 68% of anything that is normally distributed falls within one standard deviation of the mean. That means that 68% of IQ scores fall between 85 and 115 (the mean being 100 and standard deviation being 15) AND 68% of adult male heights fall between 65 and 75 inches (the mean being 70 and I am estimating a standard deviation of 5). Basically, even though the means and standard deviations change, something that is normally distributed will keep these probabilities (relative to the mean and standard deviation). By standardizing these numbers (changing the mean to 0 and the standard deviation to 1) we can use one table to find the probabilities for anything that is normally distributed.
67% as it's +/- one standard deviation from the mean
It is 0.37, approx.
mrs.sung gave a test in her trigonometry class. the scores were normally distributed with a mean of 85 and a standard deviation of 3. what percent would you expect to score between 88 and 91?
The MPG (mileage per gallon) for a mid-size car is normally distributed with a mean of 32 and a standard deviation of .8. What is the probability that the MPG for a selected mid-size car would be less than 33.2?
68 % is about one standard deviation - so there score would be between 64 and 80 (72 +/- 8)
Standard deviation is a way to describe how the data is distributed around the Arithmatic Mean. It is not a simple formula to calculate, as shown in the links.
If a normally distributed random variable X has mean m and standard deviation s, then z = (X - m)/s
The Miller Analogies Test scores have a mean of 400 and a standard deviation of 25, and are approximately normally distributed.z = ( 351.5 - 400 ) / 25 = -1.94That's about the 2.6 percentile.(Used wolframalpha.com with input Pr [x < -1.94] with x normally distributed with mean 0 and standard deviation 1.)
mean= 100 standard deviation= 15 value or x or n = 110 the formula to find the z-value = (value - mean)/standard deviation so, z = 110-100/15 = .6666666 = .6667
A normal distribution with a mean of 65 and a standard deviation of 2.5 would have 95% of the population being between 60 and 70, i.e. +/- two standard deviations.
About 98% of the population.
BMI varies from person to person and one measure of this variability is the standard deviation. Assuming the BMI is approximately normally distributed, only around 0.135% of the people will have results that are -3 sd or lower.
The Z test.
It has been determined that the wait-time for computer support is normally distributed with a mean of 30 minutes and a standard deviation of 5 min. If 100 people call, how many would you expect to wait more than 30 min? (use logic here)