### Quiz 08 Normal Curves 7.3b • Name:

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.

1. p(x ≤ 3) = __________ What is the probability a leaf blade length will be 3 cm or less?
2. p(x ≤ __________) = 0.15. Fifteen percent (15%) of cherry leaves are shorter than what length x? Determine that length x.
3. p(x ≥ __________) = 0.20. Twenty percent (20%) of cherry leaves are longer than what length x? Determine that length x.
4. __________ If a cherry bush has 900 leaves, how many leaves are as long or longer than the length in question c above?

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:

1. Bimodal
2. Normal
3. Skewed
4. Uniform
5. None of the above

[Overtime!]. I sent a note in regards the tuition increase for spring term. How much is the tuition increase for spring term?

Normal Statistics
Calculate a z value from an x z = =STANDARDIZE(x, µ, σ)
Calculate an x value from a zx = 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)