Answer the questions and shade the appropriate area under the normal curve.
How high would you have to score on the SAT so that only 2.5% of people would score better than you did if the average score is 1019 with a standard deviation of 209?
We collected data on gasoline consumption on cars made in the year 2001. If μ=21.22 mpg and σ=5.36 mpg, what percent of vehicles have a mpg rating over 25 mpg?
If your height was 2.1 standard deviations above the mean height of 68 inches and one standard deviation is 3 inches, how tall are you and what percent of people are shorter than you are?
P(x>?) = .025 or P(x<?) = 1 - .025 = .975
P(z>?) = .025 or P(z<?) = 1 - .025 = .975
[BDT1]From table A, we determine that the correct z-score is 1.96[BDT2]
Use the formula x= (σ * z) + μ
x = (209 * 1.96) + 1019
x = 1428.64
[BDT3]
P(x>25) = ?
z = σ= 5.36 = 21.22 x = 25
z = = = 0.7052 (round up to 0.71)
[BDT4]From table A, we determine that P(z<0.71) is 0.7611, however we are interested in P(z>0.71) because we want to know P(x>25). Therefore we must take (1-0.7611) to get the correct probability of 0.2389[BDT5]
P(z<2.1) = ? x = ?
To find x, simply use the formula x= (σ * z) + μ
x = (3 * 2.1) + 68
x = 74.3 inches
[BDT6]To find P(z<2.1), we use table A and determine that P(z<2.1) = 0.9821 or 98.21%[BDT7]
Pregnancies last, on average, 266 days with a standard deviation of 16 days. Between what two values does the middle one-third of pregnancies last? How many standard deviations are these values from the mean?
You want to go to graduate school to study business and must score in the 90th percentile to get into your preferred school. If GMAT scores ~ N(527,112), what must your score be in order to get into your preferred school?
The middle P(?<x<?) = 0.33
Also, the middle P(?<z<?) = 0.33
[BDT8]The picture of the normal distribution will be divided into 3 parts: the lower 1/3, the middle 1/3, and the upper 1/3.
First, use table A to determine P(z<?) = 0.33 (This is the value of z that separates the lower 1/3 and the middle 1/3). We determine that z = [-0.44]. Because the normal distribution is symmetric, we know that the upper value for z must be 0.44, so the values we are searching for will be 0.44 standard deviations away from the mean.[BDT9]
Now we can simply use our formula to calculate x
x= (σ * z) + μ
x1 = (16 * [-0.44]) + 266 x2 = (16 * 0.44) + 266
x1 = 258.96 days x2 = 273.04 days
[BDT10]
P(x<?) = 0.9 or 90% (This is the definition of a percentile)
From table A, we determine that z = 1.29 (note that there is no 0.9000 value in the body of table A, so z = 1.29 is an approximation)[BDT11]
x= (σ * z) + μ
x = (112 * 1.29) + 527
x = 671.48
A 9 ounce bag of potato chips may or may not actually contain 9 ounces of chips. If 9 ounces is actually 0.8 standard deviations below the true mean, what is the true mean given that the standard deviation of weights is 0.15? What percent of bags of chips weigh more than 9 ounces?
Z =
Z = = [-0.8][BDT12]
9 – μ = [-0.8] * 0.15
9 – μ = [-0.12]
μ = 9 + 0.12 = 9.12
[BDT13]From table A, we determine that P(z<[-0.8]) = 0.2119, but we are interested in P(x>9) so we must find P(z>[-0.8])
1 - 0.2119 = 0.7881 or 78.81%[BDT14]
[BDT1]Here, we start with the probability ABOVE some value of x and must work backward to find that value.
There are two methods to do this, both are shown here, side by side.
[BDT2]Search inside the body of table A to find the probability we need, then go outward to the row and column headings to determine the z-score.
[BDT3]Once we know the z-score, we simply use the necessary formula to calculate x.
[BDT4]Here we start with x and are asked to find a percent/probability. First we must calculate z.
[BDT5]Table A always tells us the probability BELOW a given z-score [for example P(z<0.71)], but this problem asks us to find the percent ABOVE the z-score.
We simply need to take 1-P(z<0.71) to find P(z>0.71)
[BDT6]In this case, we already know z and are asked to find x. Simply use the formula.
[BDT7]We already know z and are trying to determine P(z<2.1), so simply find the row in table A labeled “2.1” and the column “.00” and that probability is the correct answer.
Note that the row headings indicate the ones and tenths (2.10) place for the z-score, and the column headings indicate the hundredths place (2.10)
[BDT8]This is a common problem, when you are asked to find the values that bound some middle-percent of the data. Here we are asked to find the values that bound the middle 1/3 (33%) of the data. This leaves 2/3 (66.67%) of the data to the left and right of this region, 1/3 on each side.
[BDT9]When asked to find the values that bound of the middle-percent of the data, we need only find one z-score because the other z-score will have the same value but opposite sign (i.e., in this case, z is negative so the second value of z will be the same but positive).
Also, the values for z answer the second question because the value of z is equal to the number of standard deviations from the mean.
[BDT10]Remember to do calculations for both values of z because we are asked to find 2 values for x. The final answer should be written as the pair 258.96 days, 273.04 days
[BDT11]Be sure that, when you are estimating, you give a z-score that has at least 90% below it. For example, here the value of 0.9 falls between 1.28 and 1.29. I chose to use 1.29 because its probability is 0.9015, whereas 1.28 has a probability of 0.8997. A z-score of 1.28 is closer to the correct value, but it does not have the full 90% below it.
This is called being conservative, and this idea will come up more in later chapters.
[BDT12]Note here that Z=[-0.8]
[BDT13]This type of problem is uncommon, but I included it to show another type of problem that COULD come up.
[BDT14]Read through the problem again, paying close attention to the wording. You are given many values, but the questions clearly state what you must find. Be sure that you interpreted the problem correctly. If you have questions about the wording and/or meaning, please feel free to e-mail me through the SI website.