Question

Given a normal distribution with 101 and 4, and given you select a sample of n = 4, complete parts (a) through (d). Click here to view page 1 of the cumulative standardized normal distribution table Click here to view Rage 2 of the cumulative standardized normal distribution table. a. What is the probability that X is less than 95? P(X<95) = .0014 (Type an integer or decimal rounded to four decimal places as needed.) b. What is the probability that X is between 95 and 96.5? P/95< X < 96.5) = .0108 (Type an integer or decimal rounded to four decimal places as needed.) c. What is the probability that is above 102? PIX> 102) = 3085 (Type an integer or decimal rounded to four decimal places as needed.) d. There is a 70% chance that X is above what value? X= (Type an integer or decimal rounded to two decimal places as needed.)

Answer 1

XNN/LOL, 4 7 ~ NILUL, I = NILOL, PLX 295) PixLoL cgs-sol 1 9 -= p (26-3) =) 0.0013] Ploscł <g0.5 = -101 < 9s. - Lolc 96-S-LOL 2 : 2 LOL : = Plzc.-4.5) - P(124-3) o, Oto9 a plă Log) = t - Plze Log- Lola - P ( 241 ) = 10.3085). la ... PLX >20 =0.70 1-pc (20.0.70 +3-P ( X-dot ca-101 )= 0.70 E P12CX-10L ) - 0.30 = 20.30 2.241

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