#1 Pants possible: i. Total attempts: 2
Consider the Standard Kamal dtsmbuiaxi
The mean is always and the standard dcviMkm is always.
*2 Ports possible: i. Total attempts: 2
If Z is * random variable from a standard normal distnbutkxi and if I¥Z c) w 0.2. then PiZ -cl« 0.2
O True
O False
3 Points possible: 1. “foul attempts: 2
Assume Z is a random variable with a standard normal distribution ®sd d is a positive number If P(Z 4) = 0,2, then
P( - d Z d) = 0.6.
OTrue
O False
44 Posits possible: 1. Total attempts: 2
If a distribution is oonnaE then it is not possible to randomly select a value that is more than 4 standard devhriocs from the mean.
OTrue
? False
15 Points possible: 1. Total attempts: 2
Which of the following statements are TRUE about the normal distribution9 Check all that apply.
? The area to the left of a z-scorr phis the area to the right of that same z-score will always equal 1.
O The mean corresponds to the z-score of I.
Q A data value with z-score * -1.5 is located L5 standard deviations below the mean.
C The Empirical Rule only applies when a value is exactly 1,2, or 3 standard deviatiocs away from the mean.
O A z-score is the mmsbar of standard dev’tKians a specific data vahie is from the mean of the distriburkai
*6 Paints possible: 1. Total attempts: 2
Fill in the blanks.
The depth of the snow in my yard is normally distributed, with a mean of 15 inches and a standard deviation of 0.25 inches
What value is one standard deviation below the mean? inches
What value is one standard deviation above the mean?inches
What is the probability that a randomly chosen location will have a snow depth between 2 25 and 175 mcW? .... ... percent
#7 Points possible: 1. Total attempts: 2
The graph illustrates a normal distribution for the prices paid for a particular model of HD television. The mean price paid u $1800 and the standard deviation is $140,
Distribution of Prices
What is the approximate percentage of buyers who paid between $1800 and $1940? ____________________%
What is the approximate percentage of buyers who paid between $1800 and $2080? _____________________%
What is the approximate percentage of buyers who paid more than $2220? _____________________%
What is the approximate percentage of buyers who paid between $1660 and $1940?
_____________________%
What is the approximate percentage of buyers who paid more than $2080?
____________________%
What is the approximate percentage of buyers who paid between $1800 and $2220?
____________________%
#8 Points possible: 1. Total attempts: 2
A set of exam scores is normally distributed with a mean *= 84 and standard deviation » 6. Use the Empirical Rule to complete the following sentences.
68% of the scores are between and.
95% of the scores are between and.
99 .7% of the scores are between and
Joints posstrfe* 1 Total attempts: 2
In a norma! distribution. a data value located 1.5 standard deviations below the mean has Standard Score: z
In a normal distribution, a data value located 1.8 standard deviations above the mean has Standard Score, z =
In a normal distribution. the mean has Standard Score: z *
#10 Points possible: 1. Total attempts: 2
A variable z is normally distributed with mean 18 and standard deviation 7.
Round vour answers to the nearest hundredth as needed.
a) Determine the x -score for x = 22 .
x = _____________
b) Determine the z -score for x = 13.
t ~ _____________
c) What value of z has a z -score of 0.86 ?
z = _____________
d) What value of x has a z -score of 0.1 ?
x = _____________
ej What value of x has a z -score of 0 ?
X =£ ____________
#11 Points possible: 1. Total attempts: 2
Suppose that the weight of seedless watermelons is normally distributed with mean 6.4 kg. and standard deviation 1.2 kg. Let X be the weight of a randomly’ selected seedless watermelon. Round all answers to 4 decimal places where possible.
a. What is the distribution of X? X ~ N(,)
b. UTraa is the median seedless watermelon weight?kg,
c. What is the Z-score for a seedless watermelon weighing 7.8 kg?
d. What is the probability that a randomly selected watermelon will weigh more than 5.7 kg?
e. What b the probability that a randomly selected seedless watermelon will weigh between 7.1 and 7.9 kg?
f. The 75th percentile for the weight of seedless watermelons is kg.
»u romu possuxe: i. tout attempts: 4
The number of ants per acre in the forest is normally distributed with mean 42,000 and standard deviation 12,316. Let X * number of mu in a randomly selected acre of the forest, Round all answers to 4 decimal places where possible.
ft What is the distribution of X? X - N( ,)
b. Find the probability that a randomly selected acre in the forest has fewer than 45»3O9 ants.
c. Find the probability that a randomly selected acre has between 27.262 and 35.528 ants.
d. Find the first quartile.ants (round your answer to a whole number)
#13 Points possible: 1. Total attempts: 2
The average THC content of marijuana sold on the street is 10,4%. Suppose (he THC content is normally distributed with standard deviation of 1%. Let X be the THC content for a randomly selected bag of marijuana that is sold on the street. Round all answers to 4 decimal places where possible.
ft What is the distribution of X? X - N( )
b. Find the probability that a randomly selected bag of marijuana sold on the street will have a THC content greater than 10.8.
c. Find the 66th percentile for this distribution.%
#14 Points possible: 1. Total attempts: 2
T- SO skxthxs dixricu averaged 2168 votes per district for President Clinton. The standard
deviation was 599. (There are only 40 election districts in Alaska.) The distribution of the votes per district for President Clinton was bell-shaped. Let X ~ number of votes for President Clinton for an election district. (Source: The World Almanac and Book of Facts) Round all answers except part e, to 4 decimal places.
a. What is the distribution of X? X - N(,)
b. Is 2168 a population mean or a sample mean? [Setect an answer
c. Find the probability that a randomly selected district had fewer than 2199 votes for President Clinton.
d. Find the probability that a randomly selected district had between 2126 and 2260 votes for President Clinton.
e. Find the third quartile for votes for President Clinton. Round your answer to the nearest whole number.
#15 Points possible: 1. Total attempts: 2
Ixjs Angeles workers have an average commute of 28 minutes. Suppose the LA commute time is normally distributed with a standard deviation of 14 minutes Let X represent the commute time for a randomly selected LA worker. Round all answers to 4 decimal places where possible.
ft WhM is the distribution of X? X ~ N( , j
b. Fmd the probability that a randomly selected LA worker has a commute that is longer than 32 minutes.
c. Find the 85th percemile for the commute lime of LA workers. minutes
Ss» rrse; s •s't ucb cars gr c&»treerrT-’ ® tx?e?sxh essse SC «& 2^4 anM devise W a&es
W toyr Us X fee ’ar .speed *x a mx®) se^KXed ur. JU«d Ml «H*ers to 4 dec enaJ pace* where po^Ue.
a Ife’W a Se ^3Sr^atK» erf X* X * M.
t !f3mara33X€5 jhnsex Suffer jrdbMtm ±M s a wrrksg nxrt ±aa ?9eph
c tf one grfthe cars « *»ijSL.S cacwra, fax} the prab»Kbty tha M »tmek*-g between X2 wd 85 cspfe
tF%«rf rfBarilriMdM oak* ixs. cm the freeway a^it
t*T fesrss posEfcMr 1, tea 2
The rsernse erf S^se &e xx snead * Gtcmt Sc« Sprags is aoraalh &£nbt&£d * i± a mean of T? n^Mes asd a saadJrc ar*-jES3E M Sb Ezracr* Scrrose ose peraoe m the ha sprags k riBdedy caosee. Les X = me sexmhM erf line ?« persos spes s Gcwer Hut S pnoo a£ awen w 4 dcozaaJ places where poss^e,
a IWstrtatcioirfrx*^?
a F©£ tar prrsc-Cry a raexxsh seissed ptrscc ae ^e het sprop says k®ger Thro 98 mhaaer.
c tir wt semee s caEsagrrs? Jt3erw« a ttssccrat w me T%o? oar patrons ww spead ttr tea* ta?sc » or am T^f5 a rrsss: amneg ef ^?sae a preroc oe spead M tx bat sprags and sml recene she drscxxaas?ssEGEes.
FM ar 5SK7 Oar^je SLaege ’ K?& * ^r asar spcM M te feet sprass.
Qi.________
Q srara&s.
_______arPF&es
#tf ®iort5 jCEztse: t, Ixm 2
The rwmtee A^cnaK ieoie 13 gnKss erf scxSasa each diy Swoose thM the so&ra cocsyractofi rf Araenzaa aea is
-«¦ a s&K^ard desixMn of gram. Ss^pose » AstericaQ man b raadomh chosea. Let X = the aaowM rf
Gamene «^wers lo 4 dccss^ places where po&sibie.
a aaf » ^sz^axsx of X* X * bi,)
h Fad the fat ^35 .AsaerxaE ssas ccessaa beeween 9 md 9.9 gr®» rfsodars per hry.
c. Pm ws^W 18% M .Is&moE hwe c©E5®rae brrwecs whM two weigbts rf »dn»B?
_________
i. tuiai attempts: 4
The axemge ft’^nber of words in a romance novel is 61.035 and the standard deviation is 17.102, Assume the distribution is normal Let X be Uh: number of word* in a randomly selected romance novel, Round all answers to 4 decimal places where possible.
at V^hat is the distribution of X? X * N{ ,)
b. Find die proportion of all novels that are between 55384 and 67.565 words
c. (he 95th percentile for noveh is words. (Round to the nearest word)
d. The middle 80% of romance novels have from _____ words to words. (Round to the nearest word)
#20 Points possible: 1. Total attempts: 2
The average student loan debt for college graduates is $25,900. Suppose that that distribution is normal and that the standard deviation is $11,100. Let X * the student loan debt of a randomly selected college graduate. Round all probabilities to 1 decima? places and all dollar answers to the nearest dollar.
a. What is the distribution of X? X ~ N(,)
b Find the probability that the college graduate has between $ 11.550 and $26,900 in student loan debt.
c. 1 he middle 20% of college graduates’ loan debt hes between what two numbers?
Low. $____________
High: $___________
GET ANSWERS / LIVE CHAT