A, -2 0 2 B, 106 220 334 C 198 220 241.1 D 218 220 222

think these number are on the X axis that are the bell shape of the graph the 3 point

The values -2, 0, and 2 would most likely be z-values though. Unless the axis of the graph is marked as something like “z-value”, then this choice is not likely to be correct.

A typical layout though would label the mean in the center, and the values that are 2 standard deviations above and below the mean. With a mean of 220, and a standard deviation of the mean of 10.58, the values that are 2 standard deviations away from the mean are:

220 – 2(10.58) = 198.84

220 + 2(10.58) = 241.16

These values match the values given in Choice C.

use the normal distribution of SAT critical reading scores for which the mean is 510510 and the standard deviation is 114114. Assume the variable x is normally distributed.

left parenthesis a right parenthesis(a)

What percent of the SAT verbal scores are less than 550550?

left parenthesis b right parenthesis(b)

If 1000 SAT verbal scores are randomly selected, about how many would you expect to be greater than 525525?

left parenthesis a right parenthesis(a) Approximately

nothing% of the SAT verbal scores are less than 550550.

(Round to two decimal places as needed.)

a) z = (x – mu) / sigma = (550 – 510) / 114 = 0.3509

P(x < 550) = P(z < 0.3509) = 0.6372

b) z = (525 – 510) / 114 = 0.1316

P(x > 525) = P(z > 0.1316) = 0.44765

n = 1000 * 0.44765 = 447.65

use the normal distribution of SAT critical reading scores for which the mean is 505505 and the standard deviation is 112112. Assume the variable x is normally distributed.

left parenthesis a right parenthesis(a)

What percent of the SAT verbal scores are less than 675?

left parenthesis b right parenthesis(b)

If 1000 SAT verbal scores are randomly selected, about how many would you expect to be greater than 550?

left parenthesis a right parenthesis(a) Approximately

nothing% of the SAT verbal scores are less than 675.

(Round to two decimal places as needed.)

a) z = (675 – 505) / 112 = 1.5179

P(x < 675) = P(z < 1.5179) = 0.9355 = 93.55%

b) z = (550 – 505) / 112 = 0.4018

P(x > 550) = P(z > 0.4018) = 0.343915

n = 1000 * 0.343915 ≈ 343.92

A doctor wants to estimate the HDL cholesterol of all 20- to 29-year-old females. How many subjects are needed to estimate the HDL cholesterol within 33 points with 99 %99% confidence assuming sigma equals 11.7 question mark σ=11.7? Suppose the doctor would be content with 95 %95% confidence. How does the decrease in confidence affect the sample size required?

A 99% confidence level requires

nothing subjects.

(Round up to the nearest whole number as needed.)

n = (2.576 * 11.7 / 3)^2 = 101

A doctor wants to estimate the HDL cholesterol of all 20- to 29-year-old females. How many subjects are needed to estimate the HDL cholesterol within 44 points with 99 %99% confidence assuming sigma equals 13.7 question mark σ=13.7? Suppose the doctor would be content with 90 %90% confidence. How does the decrease in confidence affect the sample size required?

A 99% confidence level requires

n = (2.576 * 13.7 / 4)^2 = 78

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