The army reports that the distribution of waist sizes among female soldiers is approximately normal, with a mean of 28.4 inches and a standard deviation of 1.2 inches. Part A: A female soldier whose waist is 26.1 inches is at what percentile? Explain your reasoning and justify your work mathematically. Part B: The army uniform supplier regularly stocks uniform pants between sizes 24 and 32. Anyone with a waist circumference outside that interval requires a customized order. Describe what this interval looks like if displayed visually. What percent of female soldiers requires custom uniform pants? Show your work and justify your reasoning mathematically.

Step-by-step explanation:

1)

Percentile is related to the area under the standard normal curve to the LEFT of a certain data value (which in this case would be 26.1 inches).

On my Texas Instruments TI-83 Plus calculator, I found this area as follows:

normcdf(-100, 26.1, 28.4,1.2), where the range -100 to 26.1 represents the area (as a decimal fraction) to the left of 26.1 inches.  My result was 0.028, which corresponds to the 3rd percentile (0.028 rounds off to 0.03, which would be 3rd percentile).

2)  The mean waist size is 28.4 inches, represented by a vertical line through the standard normal curve lying between 24 and 32.  We use the same function on the calculator:  normcdf(24, 32, 28.4, 1.2).

The result is 0.9985.  Subtracting this from 1.0000, we get 0.001, or 0.1%, which is the percentage of female soldiers requiring custom uniforms.

make sure you have version 2.50

try looking in the very back of the book : )

step-by-step explanation:

you can download it here

2.2°  

step-by-step explanation:

let θ = the angle of elevation.

tanθ =77/2020

tanθ = 0.038 12

      θ = tan⁻¹0.038 12

      θ = 2.2°