Question

anner is normal with mean value wishes to establish a weight Problem 4 The weight distribution of parcels sent in a certain m 12 Ib and standard deviation 3.5 lb. The parcel service value C 17.76 ib beyond which there will be a surcharse will exceed Find the probabiliy that the weight of a randomly selecte 17.76 1b t least 2 o b) If 5 parcels are randomly selected, what is the probability that a them will be subject to surcharge? For full credit show all steps!

Answer 1

A P(X > 17.76) Normal Distribution, μ-12, σ-3.5, we convert this to standard normal using 17.76 (12) 3.5 1.6457141.65 P(Z1.65) Area to the right of 1.65 0.0495 1.65 P(X > 17.76) = P(Z > 1.65)-1-P(Z < 1.65)-1-0.9505 -0.0495 P(Z 1.65): in a z-table having area to the left of z, locate 1.6 in the left most column Move across the row to the right under column 0.05 and get value 0.9505

P(X > 17.76) = 0.0495 = 4.95% P(X 2), n = 5, p = 0.0495 P(X 2 2) 1- P(X < 2) We first calculate: P(X = 0) + P(X-1) (because these are disjoint events) P(X = 0)-() (0.0495)" (1-0.0495)5-1(0.0495)"(0.9505)5 0.775819 P(X = 1)-(!) (0.0495)1 (1-0.0495)4-5(0.0495)1 (0.9505)4 0.202015 P(X 0)+P(X 1) 0.775819+0.202015 0.9778340.978 P(X < 2) = 0.978 P( 2) = 1-P(X < 2) 1-0.978-0.022 P(X > 2) = 0.022 Using excel function 1 - BinomDist(1,5,0.0495, true) or TI-83/84 function 1 -binomcdf (5,0.0495,1), exact answer is 0.0221656196 oThank yous