Purchase Solution

# 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.

##### 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
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"
z value corresponding to ...

##### 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.