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**Unformatted text preview: **W11: :2 mama;
/\/v\ 1. (25 pts.) Human pregnancy duration is well known to be approximately normally distributed with ,u =
266 days with 0' = 16 days. a. What is the chance a pregnancy lasts 250 days or longer? .- 0—266'
may»): “22 L—J /-6 = m 2 2 a)
I-— fC—t) H r. I— .1337 b. What is the 75th percentile for human pregnancies (i.e. the time at which only 25% of pregnancies last beyond)? W 2. (20 pts.) In the game of Chuck-A-Luck you, the contestant, pick a number from 1 through 6. Then the
operator of the game throws three dice. For a $1 bet, if all three match the number you chose, you have net
winnings of $3 from the operator. If there are two matches you net $2, if there is one match you net $1, and
if there are no matches you net «$1. For a single bet of $1 the net winnings (from Exam II) have a mean of ,u E -$0.0787 and a standard
deviation of 0' E $1.113. a. Estimate, using the Central Limit Theorem, the chance of at least breaking even in 225 consecutive bets
of $1 on Chuck-A—Luck. FCZV“ "*2125 2- °> £551 04:9 b. Estimate, using the Central Limit Theorem, the chance of winning between $10 and $20 in 225
Consecutive bets of $1 on Chuck-A-Luck. U57“? s‘lhﬂ’lr mwh7na’ F(/9< 24“"4’1245‘ 20) 3. (25 pts.) A short cable is made of 40 strands of a ﬁxed length. Individually, the strands have breaking
strengths with mean 35 lb. and standard deviation 5 1b. Suppose further that the breaking strength of a cable
is roughly the sum of the strengths of the strands that make it up'. _ W Y 4 ~ - - 0- 3.7
a. Estimate the mean breaking strength for such cables. ’ + 1.1“ +9 ECY'J’ "‘4" 2”) 5 EC?.)4’ ﬂY1)+"'+’ Eé29a) = 4a (351») =® b. Estimate the standard deviation of the breaking strength for such cables. ---.-__ —--' ”*Cz*"’* 2*» )0? V0 CZ.) + "HI" Val-(2.?!) '- 4D(5Ib.)z
‘2 5+4 Jamar}; bar-«him, 5414.7“ c. Estimate the chance that such a (randomly selected) cable will support a load of 1,450 lbs. (Note: To
support a load of at least 1,450 lbs. the breaking strength must be above 1,450 lbs.) P< zit-"~1- Zfo 2. [,7'50) -. —- I‘H’o
Hz? 4» : Ni 2 34.2.5) ._ M Z a _, 34.2.5— 35 , ‘(u/r WK — {/4 $5 \é’ V('2~z./,;7) 1 This is roughly true for “short” cables. For an entertaining discussion of this as a key element of a story plot read Eyes
of the Dragon by Stephen King (a good read!). 4. Former IEng 362 student Everett Hix collected the following interarrival time data on vehicles using the
drive-up service at the Rapid City Hardees at 5‘“ Street and St. Joseph: Interarrivals (seconds) 44 156 265 25 61 15 134 34 262 187
28 15 12 85 3 7 23 19 50 97
36 128 49 91 83 68 76 6 9 44
14 31 170 15 50 29 71 55 20 19
73 32 108 37 5 78 145 6 243 For this data some computation gives 3 = 1.13 minutes and s = 1.12 minutes. Hardees, Fifth & St. Joseph, Drive-Up Interarn'vals
October 26, 2001 12pm - 1pm
Everett Hix o 50 100 150 200 50 300 350
lnterarrivals (seconds) a. (5 pts.) The above data may be ﬁt by an exponential density of the form le'" x>0 / f(x)={ o x_<.0 for some value of 2.. Give me a good estimate of A (suggestion: work in units of minutes). A"?
€16. ,L_—-— ,A—VC A J—.. 1 c: -
’3 (AHA: % ’ (:7;- " .375 «701‘ continued \\ ’Technical detail
time for which n to the Hardees in question. What is the chance that the next
at is, what is the chance that the next interarrival time is W
2
3 -.8?,¢7c
m)“ a .me. Ax.
'L 2.
<7?» 3
c —-e 1
- 9‘ ~39!!!-
¢ _e .9 5(5) 4. e )
“-5 <07o3+ .ma
‘3 to the drive-up~(note that this corresponds to 36 interarrival
ve within 50 minutes? function of time — suppose that the question concerns a pen'od of ...

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- Summer '04
- JOHNSON
- Statistics, Probability