Question

points PObStat5 7.E.078, A machine that cuts corks for wine bottles operates in such a way that the distribution of the diameter of the corks produced is well that is too small leaks and causes the wine to deteriorate; a cork that is too large doesn't fit in the bottle.) What proportion of corks produced by this machine are defective? You may need to use the appropriate table in Appendix A to answer this question. Need Help? Read ItTalk to a Tutor 7.1 points PODStat5 7.E.080 distribution that is wel asoline tank for a certain car is designed to hold 17.0 gal of gas. Suppose that the variable x - actual capacity of a randomly selected tank has a (a) What is the probability that a randomly selected tank ill hold at most 16.8 gal? P[x s 16.8) (b) What is the probability that a randomly selected tank vwill hold between 16.4 and 17.3 gal? P(16.4 s x s 17.3) (c) If two such tanks are independently selected, what is the probability that both hold at most 17 gal? P(x s 17) You may need to use the appropriate table in Appendix A to answer this question Need Help? ReadIt Talk to a Tuter

#6
a machine that cuts corks for wine bottles operates in such a way that the distribution of the diameter of the corks produced its well approximated by a normal distribution with mean 3 cm and standard deviation 0.1 CM the specifications call for corks with diameter is between 2.9 and 3.1 CM a cork not meeting the specifications is considered defective a cork that is too small leaks and causes the wine to deteriorate a cork that is too large doesn't fit in the bottle What proportion of corks produced by the machine or defective round the answer to four decimal places.

#7 a gasoline tank for a certain car is designed to hold 17.0 gallons of gas suppose that the variable x equals actual capacity of a randomly-selected tank has a distribution that is well approximated by a normal curve with mean 17 gal and standard deviation 0.2 gal round all answers to four decimal places.

please look to photo for questions to these listed below, thank you.
(a)
(b)
(c)

Answer 1

7. X~actual cap. of tank

Mean = 17

Stdev = .2

a. P(X<=16.8) = P(Z<= 16.8-17 / .2) = P(Z<=-1) = .16

b. P(16.4<=X<=17.3) = P(16.4-17/.2 <=Z<= 17.3-17/.2) = P(-3<Z<1.5) = 0.9332-0.0014=0.9318

c. P(Xbar<=17) = P(Z<=17-17/(.2/sqrt(2)) = P(Z<=0) = .50