Question

Question 11 A manufacturer knows that their items have a normally distributed length, with a mean of 15.6 inches, and standard deviation of 4.7 inches. If 23 items are chosen at random, what is the probability that their mean length is less than 18.1 inches? Pa < 18.1) = Submit Question Question 12 BO A manufacturer knows that their items have a normally distributed lifespan, with a mean of 9.3 years, and standard deviation of 2.7 years. If you randomly purchase 8 items, what is the probability that their mean life will be longer than 9 years? PE > 9) = Submit Question Question 13 BO A particular fruit's weights are normally distributed, with a mean of 207 grams and a standard deviation of 21 grams. If you pick 16 fruit at random, what is the probability that their mean weight will be between 208 grams and 213 grams P(208 < <213) Submit Question

Answer 1

Question 1] Here , Let Х be length of of items (in inches) X N N C = 15.6 02= (4.7) 2) Also, 0=25 want to calculate , probability that their meen length is less than 18.1 inches. ie. PCX 21801) we Herse, PLX X N( u = 15.6 02.(4.71°) 6 el = 15.6 NE (4•702 25 = x - 4 N Noel) olin < 9 (2< 18.1-15-6 CINN(0,1) P(X<18. 1) 18.1-es olun 18.1-15-6 2-71dar = P(Z22-6596) - 1-P (272-6596) 1- 0.0039 .9961 IPCTL 18.1) = 0.996) Question 12) X be items Let Here, lifespan of X NN/ & M = 9.3 7 (in years) 6²3 (2+72²) (2.5)) Here, n=8 X NN خر U [ o g. 3 ナ X - el olun

med life we have to calculate, poobability of their will be longer than g years .ie .pcx>g), .P( 89) P(x-4 5 Wo 7 ("INN(0,1) 2.7158 = P(23-0-3143) = P(2<0.3143) Prz>0.3143) 0.3767 PCX>g) = 0.6233 question 137 Let X be weights of fruits Here, x N NC 4 = 207, o²=(21) 2) cin groems) Here, no 16 No e = 207 (21² 2100) n ::2=&re ~ NCOI). we want to calculate, probability that theio mean weight will be between eos groems and 213 goyms. je. P(208(2013) 9 ( 208-4 <-4 < 213-4 olin 208 -201 <2L213 - 207 21176 211576 = P(0.1995<2< 1.1428) PE

: P(2082 X < 213) = P(Z < 1.1428) - P(Z 20.1905) - I - P(Z71.1428 - 1 + p (270.1905) P (270.1905) - P (27111428) a. 4245-0.1266 0.2979 . 1 P 1208 < X (213) = 0.2979 X be Question 14) Let weights of frouts Lin god mns) here XN N( = 269 02 (31) 2 Here, n=31 We know that, if XNN(-4, 6²) then Innle, &lon) and za 7-e Clon N Nooit considee P(Z > Zascore 0.00 .. Z-score= 1.55J. :: 1.555 = 차시 Elu5 .: 1.555 7-269 31 11 y = 1.555 X 31 131 +269 x = 277.65-79 grams

be weights of cows Cih X NNC l = 865, 82=132) 2) 4= 365, Questions Let y bere bere u Nf u Q. (sopa) 2= x - 4 NNCO requised hence, P ( X >400) - probability PC 2 400- Plzz p. z 400-365 3217 P (27 2.8938) 0.0019, Pix > 400) = 0.0019 Question loy let x be length of unemplo yonent (in week) X ~ Ncello 38.3 Des (F26)?) De 223 2 X u = 38.3 El Nous (7.403 223 -2=X - 4 бъ N N(0,1) то find probability that a single randomly selected value is between 36.7 and 39.7.

PC36.92X < 39.7) =P / 3607 – 38-3 39.7-38.3 <X-38.3 7.4 704 -0.2162 ez < 0.18982) . Plz 20:1892) - P(Z <-0.2162) 1- P(Z30:1892) - P(270.2162) 10. 4250 -0.4144 0.1606 :1P (86.78X (39.7) -0.1606 1. То Rind probability that is randomly selected with and 39.7 sample of size n=223 medn between 36-7 P(35.767489-3) = 1/36.9-30.3 <3723823 < 39.7-38.3 7.6/5223 7.415223 - P p ( -3.2253 -3.2288 cz (2-8282 = P(Z 2 -7 28 2 ) - P ( 2 / -3.42.8 3 ) Ĉ 1- PC Z72.8252) - P(z? 3,22887 0.0024 0.0006 - 0.9970, jec 36 GP (36.70x<39.7) = 0.9970 cost aruto insurance Question 17) Let x be annual cin dollars). here, X N N / 4 = 1031 n=59. 82=(20722)

Ť N Nr 4 = 1031 82 + Neu 207) 59 2) NCO, I). TO find P ( X <970) X-103) 20 <970 - 103 207 Zc -0.2947) Leon P1270.2947) 0.3841 lpcx < 970) 0.884) to find, P ( X < 970) e970-103) 11 Pr X-803) 207 2071559 P ( 22-2.82635) P ( 27 2.2635) 0.0118 1. P (X 970) = 0.0118