Question

During the spring, many rivers are monitored very carefully in case residents need to be warned about an impending flood. The depth (in feet) of the Susquehanna River at the Bloomsburg Bridge is measured and reported daily. In a random sample of depths, the mean 16.7, s=2.1, and shape of distribution is approxiamtely normal

Please Solve using empirical method:

a. Approximately what proportion of depths are between 14.6 and 23 ft? [ANSWER 0.835]

Part II: PLS Solve using permutation or combination method

In preparation for the coming season, a bass fisherman decides to buy 5 random lures out of the 10 new ones in the local tackle shop.

A. How many different collections of 5 new lures are possible?

B. Suppose 1 of the 10 lures is a Crazy Crawler. What is the probability that the fisherman will not select this lure?

C. Suppose 3 of the 10 are Excalibur lures. What is the probability that at least 1 of the 5 selected will be an Excalibur lure?

Answer 1

Mean (a) - 16 17 8 -- 2-1 , K 3 = 165 1-2 1 - 14 % * +34 = 16.1 + ( 3 124 1) - a3 )

99.9 95/ Ik 34%. 34%. A B (13.5% (2-35 x-38 x 28 x-8 Mean xts 4+28 xe38 50% 50/ A= 95% 5% -34% = 13.5%. B = 99.7% 34/- +13.5'7) = 21357 So; between propostion lying and x + 3s is in x-s 0.34 +0.34 +0.135 +0.0235 = 0.8385 Amy 0,8385

A 10es =252 collections There possible. B. P ( Gazy Crawler) P (The fisherman will not select this Garzy Crawler) to = 0.59049 c. Po P ( Excalibur lures) = To find; P (at least t of 5 selected is Excaliber lure) an =P(X), 1) = 1 - P(X21) =1- P(x=0) = ! می) =-{*** (*(1-2) E ) - 0.1680 7 = 0.83193. Amy 0.83193.