6.2 & 6.3 Stats HW

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A Z-Score is an area under the normal curve.

6.3.9 (#'s are left of x) Find the indicated IQ score. The graph to the right depicts IQ scores of​ adults, and those scores are normally distributed with a mean of 100 and a standard deviation of 15.

1. Use the standard normal distribution table to find the z score that has a cumulative area from the left of 0.97. If the exact area is not in the body of the​ table, avoid interpolation by selecting the closest value. If a desired value is midway between two table​ values, select the larger table value. -have to look within table at ALL .0000 #'s to find the one when rounded to only 2 decimals places =.97 z score = 1.88 2.Convert a standard normal random variable z to a normal random variable x by using the formula​ below, where μ is the mean and σ is the standard deviation. x=μ+[(z)(σ)] 100+(1.88 x 15) = 128.2 which is the indicated IQ score

What conditions would produce a negative​ z-score?

A​ z-score corresponding to an area located entirely in the left side of the curve would produce a negative​ z-score.

85th percentile.

data value associated with an area of 0.85 to its left

calculating area to left of z score

http://www.danielsoper.com/statcalc/calculator.aspx?id=2

calculating z score

http://www.socscistatistics.com/tests/ztest/zscorecalculator.aspx

​Men's heights are normally distributed with mean 68.7 in and standard deviation of 2.8 in. ​Women's heights are normally distributed with mean 63.2 in and standard deviation of 2.5 in. The standard doorway height is 80 in. a. What percentage of men are too tall to fit through a standard doorway without​ bending, and what percentage of women are too tall to fit through a standard doorway without​ bending? b. If a statistician designs a house so that all of the doorways have heights that are sufficient for all men except the tallest​ 5%, what doorway height would be​ used?

- Suppose that a random variable x is normally distributed with mean μ and standard deviation sσ. The area below the normal curve represents a proportion or probability. A. In order to determine what percentage of men are too tall to fit through the doorway without​ bending, draw a standard normal curve. Label the given values for x and μ. Then shade the appropriate region. 1. The next step is to convert 80 to its corresponding​ z score. Use the fact that z=(x−μ)/σ (80 - 68.7) / 2.8 = 4.04 Recall that the area under the normal curve is equal to 1. To find the area under the curve to the right of the​ z-score subtract the cumulative area to the left of the​ z-score from 1. - Use a normal distribution table to find the cumulative area to the left of z=4.04 =.9999 -now subtract the area 0.9999 from 1 to find the area of the shaded region. =.00010 To find the percentage of men that are taller than 80 ​in, convert 0.0001 to a percent. = .01% percentage of women who are too tall is the same because z score is 6.72 = .9999 B. - Recall that the standard normal distribution table gives the cumulative area to the left of the​ z-score. Recall that the total area under the normal curve is 1 and the area to the right of the​ z-score is 0.05. - to find area to left of z score 1 - .05 = .95 -Use a standard normal table to determine the​ z-score that corresponds to 0.95. z = 1.645 - To obtain the​ x-value use the fact that x=μ+​(z • σ​). 68.7+(1.645 x 2.8) = 73.3 ​Thus, the statistician would design a house with a doorway height of 73.3 in.

6.3.13 (drawing my own graph & when only 1 value of x is given) Assume that adults have IQ scores that are normally distributed with a mean of μ=100 and a standard deviation σ=20. Find the probability that a randomly selected adult has an IQ less than 132 (this is the x).

- Suppose that a random variable x is normally distributed with mean μ and standard deviation σ. The area below the normal curve represents a proportion or probability. -To help visualize the area of​ interest, draw a standard normal curve. Label the given values for x (randomly selected adult has an IQ less than) and μ. 1. The next step is to convert the value of x to the equivalent​ z score using z=(x-μ)/σ (132-100)/20 = 1.6 2. Use a standard normal distribution table to find the cumulative area to the left of z=1.6 P(z<1.6)= .9452 (just look at table for 1.6 z score) hus, the probability that a randomly selected adult has an IQ less than 32 is 0.9452

If this branch of the military changes the height requirements so that all women are eligible except the shortest​ 1% and the tallest​ 2%, what are the new height​ requirements?

- The z score with an area of​ 1%, or​ 0.01, to its left is -2.33 -Convert the standard normal random variable z to the normal random variable x x=μ+​(z • σ​). for maximum height - First find the z score with​ 2% of the total area under the standard normal curve to its right or ​100% - 2% =​98% of the total area under the standard normal curve to its left The z score with an area of​ 98%, or​ 0.98, to its left is 2.05 -Convert the standard normal random variable z to the normal random variable x x=μ+​(z • σ​).

​Women's heights are normally distributed with mean 63.6 in and standard deviation of 2.5 in. A social organization for tall people has a requirement that women must be at least 70 in tall. What percentage of women meet that​ requirement?

- determine that you're looking at area to the right because they must be at least 70 find z score for 70 =2.56 area to the left =.9948 we need area to right 1-.9948 = .0052 convert to %% 52%

6.3.17 Assume that adults have IQ scores that are normally distributed with a mean of 105 and a standard deviation 20. Find P15​, which is the IQ score separating the bottom 15​% from the top 85​%

-Often we are interested in calculating the value of a normal random variable required for the variable to correspond to a certain proportion or probability. - P15 means that 15​% of the data is less than the random variable x. That​ is, the area under the normal curve to the left of x is equal to 0.15. 1. Now use a standard normal table to find the​ z-score that corresponds to the shaded area. First look in the body of the table and find the value closest to 0.15 =.1492 Find the corresponding​ z-score for .1492 z= -1.04 2. To obtain the corresponding​ x value use the fact that x= μ + [(z)(σ)] 105+(-1.04 x 20) = 84.2 Thus, the IQ score that separates the bottom 15​% from the top 85​% is P15 = 84.2

6.3.19 Assume that adults have IQ scores that are normally distributed with a mean of 101 and a standard deviation of 15. Find the third quartile Q3​, which is the IQ score separating the top​ 25% from the others.

-Since​ 25% of the area under the normal curve is to the right of the third​ quartile, that means that​ 75% of the area under the normal curve is to the left of the third quartile. 1. Use the standard normal distribution table to find the z score that corresponds to an area of​ 75%, or 0.75. z score = .67 2. Convert a standard normal random variable z to a normal random variable x by using the formula​ below x= μ + [(z)(σ)] 101+(.67 x 15) =111.1 which is the 3rd quartile Q3

6.3.11 (when 3's are to right of x) Find the indicated IQ score. The graph to the right depicts IQ scores of​ adults, and those scores are normally distributed with a mean of 100 and a standard deviation of 15.

1. Note that the given area is to the right of x. Since the standard normal distribution table contains areas to the left of the corresponding z​ scores, first find the area to the left of x. The area under the normal curve is always -Find the area to the left of x by finding the cumulative area to the left of 0.80 cum. area to left of .8 = 1 - .8 = .2 2. Use the standard normal distribution table to find the z score that has a cumulative area from the left of 0.20.2. If the exact area is not in the body of the​ table, avoid interpolation by selecting the closest value. If a desired value is midway between two table​ values, select the larger table value. Find the z score that corresponds to an area of 0.2 z score = -.84 3. Convert a standard normal random variable z to a normal random variable x by using the formula​ below, where μ is the mean and σ is the standard deviation. x=μ+[(z)(σ)] x= 100+(-.84 x 15) = 87.4

The distribution of IQ scores is a nonstandard normal distribution with a mean of 100 and a standard deviation of 15. What are the value of the mean and standard deviation after all IQ scores have been standardized by converting them to z scores using z=(x-μ)/σ ?

The mean is 0 and the standard deviation is 1. Standardizing the scores to z scores makes the distribution a standard normal distribution.

(for when graph has 2 numbers) do everything for both numbers

To find the area under a curve that is between the 2 z=###'s subtract highest z score from 2nd z score

(for when graph only has 1 number) Find the area of the shaded region. The graph to the right depicts IQ scores of​ adults, and those scores are normally distributed with a mean of 100 and a standard deviation of 15. (graph has 110 indicated as shaded area(

x=110 μ=100 σ=15 (110-100)/15 = .67 zscore of x=110 is .67 The shaded region in the graph to the right is the area to the left of 0.67 in a standard normal distribution with a mean of 0 and a standard deviation of 1. look for .6 in the standard normal distribution table -- look acroos columns until you # below .07 column and still in .6 row The area to the left of z=0.67 is .7486 = area of shaded region