1. Surinam cherry (Eugenia uniflora) leaves have a mean leaf blade length µ = 3.9 centimeters with a standard deviation of σ = 0.7 cm. For this exercise assume that the leaf blade lengths are normally distributed.
2. If the shape of the distribution of data measurements x is bimodal, then the distibution of the sample means x from random samples of 30 or more leaves each will be:
[Overtime!]. I sent a note in regards the tuition increase for spring term. How much is the tuition increase for spring term?
Normal Statistics | |||
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Calculate a z value from an x | z | = | =STANDARDIZE(x, µ, σ) |
Calculate an x value from a z | x | = z σ + µ | =z*σ+µ |
Find a probability p(x) from a z value where the probability p is the area to the left of z. | =NORMSDIST(z) | ||
Find a z value from a probability p, where p is the area to the left of z | =NORMSINV(p) |
Calendar • Statistics • Lee Ling • COMFSM