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.
Yes, to approximately standard normal.If the random variable X is approximately normal with mean m and standard deviation s, then(X - m)/sis approximately standard normal.
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 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.)
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.
If a normally distributed random variable X has mean m and standard deviation s, then z = (X - m)/s
Yes, to approximately standard normal.If the random variable X is approximately normal with mean m and standard deviation s, then(X - m)/sis approximately standard normal.
The mean and standard deviation. If the data really are normally distributed, all other statistics are redundant.
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 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.)
A food processor packages orange juice in small jars. The weights of the filled jars are approximately normally distributed with a mean of 10.5 ounces and a standard deviation of 0.3 ounce. Find the proportion of all jars packaged by this process that have weights that fall below 10.875 ounces.
94.99% of an 8oz. cup
68.2%
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.
True.
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.
(X-61)/10.2=.878 X = 69.95 approximately 70
There are approximately 16.4% of students who score below 66 on the exam.