Mean and standard deviation of a normal distribution
Not what you're looking for?
In grading Oranges into "A","B" and "C", a large orchard uses weights to distinguish Oranges. Any Orange weighing more than 2 ounces is classified as grade "A", while an Orange weighing less than 0.75 ounces is classified as grade "C". If the day's pick shows 16.6% are grade "A" and 6.68% are grade "C", determine the Mean and Standard Deviation. Assume the weights are normally distributed.
Hint: First calculate the appropriate z values for the probabilities given,then set up the equations to solve for the Mean and the Standard Deviation.
Purchase this Solution
Solution Summary
Mean and standard deviation of a normal distribution are calculated using the data given.
Solution Preview
Let
M = mean
s = standard deviation
Orange weighing less than 0.75 ounces is classified as grade "C
6.68% are grade "C"
z value corresponding to probability 6.68% is -1.500056
Thus z1=(0.75-M )/s = -1.500056 --- Equation (1)
Any Orange weighing more than 2 ounces is classified as grade "A"
16.6% are grade "A"
z value corresponding to ...
Purchase this Solution
Free BrainMass Quizzes
Measures of Central Tendency
Tests knowledge of the three main measures of central tendency, including some simple calculation questions.
Terms and Definitions for Statistics
This quiz covers basic terms and definitions of statistics.
Measures of Central Tendency
This quiz evaluates the students understanding of the measures of central tendency seen in statistics. This quiz is specifically designed to incorporate the measures of central tendency as they relate to psychological research.
Know Your Statistical Concepts
Each question is a choice-summary multiple choice question that presents you with a statistical concept and then 4 numbered statements. You must decide which (if any) of the numbered statements is/are true as they relate to the statistical concept.