Question

please answer entire question :)

The price to snorkel in Maui depends on the time of year you travel to the island. The average price of a three hour snorkling trip with Maui Snorkling Charters is $169 and the standard deviation is $30. Suppose that 16 snorklers are randomly chosen throughout the year. Round all answers to 4 decimal places where possible. a. What is the distribution of I? - NG b. For the group of 16, find the probability that the average snorkling trip price is between $176 and $181. c. Find the third quartile for the average snorkling trip price for this sample size. $ (round to the nearest cent), d. For part b), is the assumption that the distribution is normal necessary? O No Yes

Answer 1

selected i.e. N=16_-1 soll: Let X: price of snorkel in Maui. Given_lhat, vindom sample_of_16_shooklers_is also LE(x) = $169 and S.D.(x) = $30 Variance 52 (30) and - goo - I- 169 16 Dishi buon 4 x Answer: Given i=1,2 XINN (169,302) We know that if Xin N14,62) then X NC 52) goo -295 here 02 Sn 4 Thus, CNN (169,225) ENN (169, 152) 2-169 15 15 15 0.6808 We need to find. PC176 < Ž < 181) - PC 146 <Ž 2181) = P(146–169 181-459) = P(0.4667 < z < 0.8000) =p (220.8) - P(220.4667) 0.7881 i From standard = 0.1073 normal prob.table. Notice that_by_Central limit_theorem, fry any qandom variable with E(X)=ll and s.d.(x)=0 then for sufficiently longe Sample, ANCO,1) We used this paperty for finding above probability PC 19 6 3 Ž< 181) = 0.1073 s/n CS Scanned with CamScanner

To find good quartile CQ3) We know for normal distribution Q = l + (0.675)*standard deviation Hence 93 = 169+ (0.675) x 15 = 179.135 Rz. 179 is not @ NO! the assumption_thet the distaibution is normal Because for any necessary x with ECxJelly rondom variable Vex) we have = 5 then for large sample of size n Z = X- ~ N(0.1) san This is known as central limit theorem. therem for finding probability the can apply this probability in CS Scanned with CamScanner