Ch. 6 Test & HW

Determine the area under the standard normal curve that lies to the left of: A. z = 0.87 B. z = -0.16 C. z = -0.64 D. z = -0.24

(Use Table II to find) A. 0.8078 B. 0.4364 C. 0.2611 D. 0.4052

Determine the area under the standard normal curve that lies between: A. z = -1.66 and z = 1.66 B. z = -1.33 and z = 0 C. z = -0.64 and z = 0.44

(do larger [usu +] value of z - smaller [usu -] value of z) A. 0.9146 0.9573 - 0.0427 B. 0.1255 0.5000 - 0.3745 C. 0.9373 0.9838 - 0.0465

Find the value of zα: z0.32

0.47 zα = z-score w/ area "α" to its RIGHT, under the standard normal curve. If 0.32 = area to the right, then 1-0.32 = 0.68 is the area to the LEFT. Use table to find z-score.

A variable is normally distributed with mean 18 and a standard deviation 2. A. Determine the quartiles of the variable. Q1 = ___ Q2 = ___ Q3 = ___ B. Obtain and interpret the 80th percentile. a. The 85th percentile is the number that divides the bottom 85% of the data from the top 15%. b. The 85th percentile is the number that occurs in the data 85% of the time. c. The 85th percentile is the number that divides the bottom 15% of the data from the top 85% of the data. C. Find the value that 65% of all possible values of the variable exceed. D. Find the 2 values that divide the area under the corresponding normal curve into a middle area of 0.95 and 2 outside areas of 0.025.

A. Q1 = 16.66 Q2 = 18 Q3 = 19.34 B. 20.08 a. The 85th percentile is the number that divides the bottom 85% of the data from the top 15%. C. 17.22 D. 14.08 and 21.92

Assume that the variable under consideration has a density curve. It's given that 26.1% of all possible observations of the variable are less than 15. A. Determine the area under the density curve that lies to the left of 15. B. Determine the area under the density curve that lies to the right of 15.

A. 0.261 26.1% / 100% =... B. 0.739 1-0.261

A variable is normally distributed with mean 16 and standard deviation 4. A. Find the % of all possible values of the variable that lie b/w 14-21. B. Find the % of all possible values of the variable that exceed 8. C. Find the % of all possible values of the variable that are less than 15.

A. 58.59% Convert to standardized normal variables: use z = (x-µ)/σ = (14-16)/4 = -0.5 and = (21-16)/4= 1.25 Find areas for these values. -0.5 = 0.3085 1.25 = 0.8944 Take bigger - smaller, to find area between the values. 0.8944 - 0.3085 = 0.5859 x 100% = 58.59% B. 97.72% Convert to standardized normal variable: z = (8-16)/4 = -2 Find area: 0.0228 This is area to the LEFT of -2; b/c we're looking for area to the RIGHT (original question asks "exceeds 8") do total area (1) - 0.0228 = 0.9772 x 100% = 97.72% C. 40.13% Convert to standardized normal variable: z = (15-16)/4= -0.25 Find area: 0.4013 x 100% = 40.13%

The figure shows a relative-frequency histogram for the heights of female students. Height (in) | Freq. | Rel. freq. 60<61 | 2 | 0.0065 61<62 | 7 | 0.0227 62<63 | 29 | 0.0939 63<64 | 74 | 0.2395 64<65 | 97 | 0.3139 65<66 | 67 | 0.2168 66<67 | 25 | 0.0809 67<68 | 6 | 0.0194 68<69 | 2 | 0.0065 -------------------------------------------- 309 | 1.0000 A. The area under the normal curve w/ parameters u=64.4 and (sigma)=1.7 that lies to the left of 65 is 0.6368. Use this info to estimate the percent of female students who are shorter than 65". B. Use the relative-frequency distribution to obtain the exact percentage of female students who are shorter than 65".

A. 63.68% -- Draw curve: µ = 64.4, σ=2, points on outer graph = 62.7 and 66.1. According to table, area for selected region = 0.6368.x100% =... B. 67.65% -- Add all of the relative frequency values for 60≤61 through 64≤65 then x100.

How is a normal probability plot used to detect outliers? a. All observations are used to construct the normal probability plot, and any observations that fall well outside the overall pattern of the data may be outliers. b. All observations are used to construct the normal probability plot, and any observations that have a normal score greater than 2 in magnitude may be outliers. c. All observations are used to construct the normal probability plot, and any observations that do not fall on a straight line may be outliers. d. All observations are used to construct the normal probability plot, and any observations above 0 may be outliers.

a. All observations are used to construct the normal probability plot, and any observations that fall well outside the overall pattern of the data may be outliers.

A curve has area 0.612 to the left of 65 and area 0.338 to the right of 65. Could this curve be a density curve for some variable? Explain your answer. a. The curve couldn't be a density curve b/c the total area under the curve is less than 1. b. The curve couldn't be a density curve b/c the total area under the curve is equal to 1. c. The curve could be a density curve b/c the total area under the curve is greater than 1. d. There's insufficient info to determine if this curve could be a density curve for some variable.

a. The curve couldn't be a density curve b/c the total area under the curve is less than 1.

With which normal distribution is the standard normal curve associated? a. The normal distribution with a mean of u and a standard deviation of σ. b. The normal distribution with a mean of 0 and a standard deviation of 1. c. The normal distribution with a mean of x and a standard deviation of s. d. The normal distribution with a mean of 1 and a standard deviation of 0.

a. The normal distribution with a mean of u and a standard deviation of σ.

A variable is approximately normally distributed. If you draw a histogram of the distribution of the variable, roughly what shape will it have? a. The histogram of the distribution of the variable would depend on the values of the data. b. The histogram of the distribution of the variable would be roughly bell shaped. c. The histogram of the distribution of the variable would have 1 peak and a long tail to the left. d. The histogram of the distribution of the variable would have 1 peak and a long tail to the right. e. There's insufficient info to determine the shape of the histogram of the distribution of the variable.

b. The histogram of the distribution of the variable would be roughly bell shaped.

Sketch the normal distribution with µ=1 and σ=2. Choose the correct graph. (all are bell-shaped) a. middle point: 2 next points: -1 5 outer points: -7, 11 b. middle: 2 next: -4, 8 outer: -7,11 c. middle: 3 next: -1,7 outer: -3,9

b. middle: 2 next: -4, 8 outer: -7,11

Under what circumstances is using a normal probability plot to assess the normality of a variable usually better than using a histogram, stem-and-leaf diagram, or dotplot? a. A normal probability plot should be used when the data are highly skewed. b. A normal probability plot should be used when there are more than 30 data values. c. A normal probability plot should be used when sample sizes are relatively small.

c. A normal probability plot should be used when sample sizes are relatively small

Use a standard normal distribution table to obtain the z-score that has an area of 0.999 to its right. z = _____

z = -3.09 z-score table gives area to the LEFT. Total area = 1. Do 1-0.999 to find out the total area to the left of the z-score you're looking for = 0.001 -3.1, -3.09 and -3.08 all have z-score of 0.001. Mean of those = -3.09 and that's the answer.