Question

section 3.5 spotlight on statistics
objective 1 - understand standard error of the mean
guided practice activity #11 - z - scores and standard error
the histograms we saw above are discrete versions of idealized continuous distributions, such as
the following normal distribution:

below + 3σ?

1. on the normal curve above what percentage of data values lie within two standard deviations

of the mean (from - 2σ to + 2σ)?

1. here are some fun questions. make an educated guess 😊:

a. 95% of us women fall between what two heights?
b. 95% of us men fall between what two heights?

Explanation:

Step1: Analyze the normal distribution curve

In a normal distribution, about 95% of the data lies within \( \mu\pm2\sigma\) (where \(\mu\) is the mean and \(\sigma\) is the standard deviation).

Step2: Apply to the problem

For the height of US women and men (assuming their heights follow a normal distribution), 95% of the data (heights) will lie within two standard deviations of the mean. But since the mean (\(\mu\)) and standard deviation (\(\sigma\)) values for US women's and men's heights are not given in the problem (common values: for US women, \(\mu = 64\) inches and \(\sigma= 2.5\) inches; for US men, \(\mu = 69\) inches and \(\sigma = 2.8\) inches). If we use these assumed values:

• For women: \(64 - 2\times2.5=59\) inches and \(64 + 2\times2.5 = 69\) inches.
• For men: \(69-2\times2.8 = 63.4\) inches and \(69+2\times2.8=74.6\) inches.

Answer:

a. If we assume \(\mu = 64\) inches and \(\sigma = 2.5\) inches for US women's heights, 95% of US women fall between \(59\) inches and \(69\) inches.
b. If we assume \(\mu = 69\) inches and \(\sigma=2.8\) inches for US men's heights, 95% of US men fall between \(63.4\) inches and \(74.6\) inches.