This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Find the indicated region under the
standard normal curve. 21=D, 2;} =1.6 Find the area that corresponds to the two z—scores. Then subtract the smaller
area from the larger area. Alternatively, you can use the normalc df function on
a graphing calculator. The area under the standard normal curve in the shaded region is
(Round to four decimal places.) Find the indicated area under the standard normal curve. To the left of: = 3.00 Choose the correct probability. Find the area to the left of the zscore using a zscore table or using the
normalcdf function on a graphing calculator.
Find the indicated area under the standard normal curve. To the right ofz = D.ED Choose the correct probability.
0 a. 0.420? C) b. 0.?645 Use the zscore table to find the area to the left ofz = D. 20. Subtract that area
from 1 to find the area to the right of the z—score. Alternatively, use the
© [1 0.5?93 normalch function on a graphing calculator. O c. n.21o4 Find the indicated arm under the standard normal curve.
Betweenz= I] andz= 0.8 Choose the approximate area. 0 a. 0.1441 Find the areas corresponding to the two 2: scores. Find their difference.
Alternatively, use the normalch function on a graphing calculator. Find the indicated area under the standard normal curve. Betweenz= —l.SS andz= 2.30 Choose the approximate probability.
0 a. mass 0.0631 Find the arms corresponding to the two 2 —scores. Then ﬁnd their difference.
Alternatively, use the normalcdf function on a graphing calculator. Find the probability of z occurring in
the indicated region. Use the Standard Normal table to ﬁnd the arm that corresponds to the zscore.
The arm is equivalent to the probability ofz occurring in the region. Alternatively, use the normalcdf function on a graphing calculator. The probability of z occurring in the shaded region = D. 3359 . (Round to four decimal places.) Find the probability of Z occurring in
the region. 2I 21 = 2.0
Find the arm that corresponds to the two z—scores. Then subtract the smaller arm from the larger area. The difference in areas is equivalent to the probability The probability ofz occurring in the shaded region is 0.8304 _ of z occuring in the region between the zscores. Alternatively, use the
(Round to four decimal places.) norrnalcdf function on a graphing calculator. Find the probability using the standard normal distribution. P(E.83 < z c: o). Find the area that corresponds to the two zscores. Then subtract the smaller
arm from the larger arm. The difference in arms is equivalent to the probability P(E.83 : z q: o) = (Round to four decimal places.) of z occuring in the region. Alternatively, use the norrnalcdf function on a
graphing calculator. Find the indicated probability using the standard normal distribution. P(l.55 <2 {1.55) Choose the correct probability.
Ci) a. ESTES 0.4395 Find the area to the left ofz= l.55 and to the left ofz = 1.55. Then subtract
the arm to the left of z = —1.55 from the arm to the left of: = 1.55.
Alternatively, use the norrnalcdf function on a graphing calculator. Assume the random variable x is normally distributed With rnmn ,u. = 82 and standard deviation
0' = 11. Find the probability. Prx < so) Pr: < 60) = (Round to four decimal places.) Find the percentage of the arm to the left of the zscore corresponding to
x= 60. 1—.“
CF . Use the formula 2 = Assume the random variable x is normally distributed with mmn it = 63.6 and and standard
deviation 0' = 2.5. Find the indicated probability. Prx > T0) Find the arm to the right of the zscore corresponding to x = 70. P(x > T0) = ooosz (Round to four decimal places.) Use the formula 2 = x—rt
J . Assume the random variable x is normally distributed with mmn it = 63.6 and standard deviation
0' = 2.5. Find the indicated probability. P(64 < x < 68) Find the area between the two zscores corresponding to x = 64 and X = 63. P(64 :1 x =1 68) = 0.39?2
(Round to four decimal places.) Use the formula Z = rest
J . Suppose the height of a population is normally distributed with a mean of 69. 2 inches and a
standard deviation of 2.9 inches. A member of this population is randomly selected. Find the
probability that the person's height is more than TEI inches. Probability that the member's height is more than TD inches = 0.3913 (Round to four decimal places.) Use azscore table to find the percentage of the area to the right of the zscore
corresponding to a height of TEI inches. Suppose that a population of adults is surveyed to measure the number of hours per week spent on
home computers. In the survey, the number ofhours is normally distributed with a mean of 7" hours and a standard deviation of 1 hour. Find the probability that a randomly selected adult
spends between 7".3 hours and 9.7" hours on a home computer per week. 130.3 < hours on home computer < 9.?) = .3?86 (Round to four decimal places.) Calculate the corresponding z—scores with the following formula. Number of hours — an Number of hours
Standard deviation z: Find the arm bounded by the two z—scores. Suppose that a population of adults is surveyed to mmsure the number of hours per week spent on
home computers. In the survey, the number of hours is normally distributed with a mean of 7 hours and a standard deviation of 1 hour. Find the percent of the adults that spend more than
7.3 hours on a home computer per week. If 2TB adults are randomly selected, about how many
would you expect to say they spend less than 6.7" hours per week on a home computer? Percent of adults who spend more than 7".3 hri‘week on a home computer = 33.21 (Round to the nmrest hundredth of a percent.) _ _ _
Calculate the corresponding zscore With the follovving formula. If 27”] adults are randomly selected, about how many would you expect to say they spend less Number of hours — Mean number of hours
than 6.? hrsi’week on a home computer? Standard deviation 103' 2 adults Find the area to the right of the zscore. Convert the area to a percent by
(Round to the nearest tenth.) multiplying by loo. First, find the z—score corresponding to 6.? hoursfweek. Find the area to the
leﬁ of the zscore under the standard normal curve. Then multiply the area by
the number of adults selected. Use the Standard Normal Table to ﬁnd the z—score that corresponds to the cumulative area. If the
arm is not in the table, use the entry closest to the area. If the arm is halfway between two entries,
use the zscore halfway between the corresponding zscores. DD? 2= (Round to the nmrest hundredth.) The arm to the left of the z—score you are trying to ﬁnd is Ill]? Using the table,
ﬁnd the arm that is closest to ELEIT; this arm corresponds to the correct
zscore. Find the indicated zscore
shown in the graph. Ara=040l3 Using the table, ﬁnd an ara that is closest to the given ara. Find the zscore Z = corresponding to that ara. Alternatively, use the invNorrn function on a
(Round to the nearest hundredth.) _ hm calculator Find the indicated zscore
shown in the graph. Ara= 0.6ﬁ23 Using the zscore table, find an area that is closest to the given area. Then find
Z = the zscore corresponding to that area. Alternatively, use the invNorm function (Round to the narest hundredth.) 011 El graphing calculator. Find the zscore that has 64. 8% of the distribution's area to its left. 2 =
Using the table, ﬁnd the arm that is closest. to 54 8% in deeirnal forrn. Then [:leund tn the nearest. ﬁncl the z—score that corresponds to that area. Alternatively, use the invNorm
function on a graphing calculator. Find the zscore that has 70% of the distribution's area between 2 and z. Remember, the symmetry of the standard normal curve says the area to the left i z = of 72 is equal to the area to the right ofz. The total area is 1 a on. Find half of that area. Then use the symmetry of the normal curve to ﬁnd the zseore with
(Round to the narest hundredth.) area D 14 to the ﬁght You sell a brand of tires that has a life expectancy that is normally distributed, with a man of
55000 miles and a standard deviation of 2500 miles. You want to give a guarantee for free
replacement of tires that don't war well. How should you word the guarantee if you are willing to
replace 10% of the tires you sell? Tires that war out by 51800. miles will be replaced free of charge. (Round to the nearest hundred miles.) Find the zscore corresponding to area 0.10 to the left of 2. Replace this score
and the man and standard deviation in the formula for z. Solve for the value, 2:.
using the formula below. A population has a man ,u. = 100 and a standard deviation 0' = 15. Find the man and the
standard deviation of a sampling distribution with the sample size n. n=240 Man of the sample = (Type an integer.) Standard deviation of the sampling distribution = . R 13 th ﬁn [1i .1: , . L131 th ula .
(Round to the nearest hundradth') ernem er e samp g stn ution man is eq to e pop tion man. The standard deviation of the sampling distribution of the sample mans, 0'3: , is
equal to the population standard deviation, 0' , divided by the square root of n. 0'}: i
v”? The prices ofa certain piece of electronic equipment are normally distributed, with a man of $355
and a standard deviation of $19. Random samples of size 55 are drawn from this population and
the man of ach sample is determined. Use the Central Limit Theorem to find the man and
standard error of the mean of the sampling distribution. Man of the sampling distribution = $
(Round to the narest dollar.) 0'
Standard error ofthe sam lin distribution = 5_.56 Recall that CI“ = ' .
p g Recallthattti=tL " 1’? (Round to the narest cent.) The per capita consumption ofa food item by people in a certain country was normally distributed
with a man of 115 lb and a standard deviation of 37".9 lb. Random samples of size 19 are drawn
from this population and the mean of each sample is determined. Use the Central Limit Theorem to
ﬁnd the mean and standard error of the mean of the sampling distribution. The mean ofthe sampling distribution = lbs
The standard error of the sampling distribution = lbs (Round to the nearest hundredth.) The standard deviation of the sampling distribution of the sample mans is
found using the following formula. The man of the sample mans is equal to the population man. 0' Suppose the population man annual salary for a certain profession is Iv. = $33300. A random
sample of 55 professionals is drawn from this population. What is the probability that the man calculate me 2.5mm for f = Minn, Find the area to 111131313, ofthe 2—score;
salary Ufthe sample, f, is less than $37799? Assume 0' = $3199. this value is equivalent to the probability. Use the following formula to ﬁnd the ZSCDI’E. Hr < arson) = o.oo34 T—F‘ (Round to four decimal places.) 2 = mixW The man height of women in a certain population (ages 2029) is .u = 66 inches. A random
sample of4U women in this age group is selected. What is the probability that f, the man for the
sample, is grater than 66.9 inches? Assume 0' = 2.?5 inches. E'ind the probability that the man height of the sample is grater than 66.9 inches.
0.0192
(Round to four decimal places.) 0'
The standard deviation of the sample mans is denoted by C"; = ﬂ . Convert the man height f = 66.9 to azscore using the formula 2 =
f— F'
o’l/‘v' 11
Don't forget that the area to the right of a zscore is equal to 1 minus the area to the leﬁ ofa zscore. Match the binomial probability with the corresponding statement. . Refer to a zscore table to find the area to the left of thez score. Hrs 53) Choose the correct statement. © 3 P(there are at most 53 successes) O b. P(there are more than 53 successes) C) c pahere are less than 53 successes) Review the interpretations for the inequality symbols. "Less than" and "grater
than" do not include the number 53. "At lastII and "at most" include the number
0 9 P(there are at last 53 successes) 51 Use a correction for continuity and choose the correct normal distribution statement for the given
binomial probability statement.
13(x > 6) Choose the correct statement. @ a. 13(x > 6.5) o b. 13(x > do) Apply a continuity correction. Since the inequality is a strict inequality, it does
0 c. P(x > 5.5) not include the probability ofx = 6. Either subtract 0.5 from the left endpoint of
0 d. 1:3 > To) this area or add 0.5 to the right endpoint ofthis area. Suppose 52% ofa certain population say that chocolate chip is their favorite cookie. You randomly
select 33 people from the population and ask each if chocolate chip is his or her favorite cookie.
Find the rela ed probabilities. Use the normal distribution to approximate the binomial distribution. Find the pro oability that exactly 20 people will say chocolate chip is their favorite.
0.0341 (Round to four decimal places.) Find the probability that at least 20 people will say chocolate chip is their favorite. To apply the Bonnan curfew0n to "exactly 20.. means you want the area D'EUM (Round to four deem places) between x = 20 minus 0.5 and x = 20 plus 0.5. Find the area under the normal curve between 19.5 and 20.5 with a mean equal to rip, a standard deviation
Find the probability that fewer than 20 people will say chocolate chip is their favorite.
equal to 1} n q . 0.?926 (Round to four decimal places.) 'At least 20' means 20 or more, so applying the continuity correction means you
want to find the area under the normal curve to the right of 19.5 with a mean USE the faCt that PCfE“WEI' than 2E0=1  F'Cnat fewer than 20)=1  Hat least equal to up, a standard deviation equal to 1} n q . 2'33" ...
View
Full
Document
This note was uploaded on 04/01/2012 for the course MGQ 301 taught by Professor Orrange during the Fall '09 term at SUNY Buffalo.
 Fall '09
 Orrange

Click to edit the document details