### 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.

- p(x ≤ 3) = __________ What is the probability a leaf blade length will be 3 cm or less?
- p(x ≤ __________) = 0.15. Fifteen percent (15%) of cherry leaves are shorter than what length x? Determine that length x.
- p(x ≥ __________) = 0.20. Twenty percent (20%) of cherry leaves are longer than what length x? Determine that length x.
- __________ 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:

- Bimodal
- Normal
- Skewed
- Uniform
- 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 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