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## A sample of n=16 observations is drawn from a normal population with µ=1000 and s=200. Find the following. i) P( >1050) ii) P(960< <1050) Part b) (3 marks) An automatic machine in a manufacturing proc

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A sample of n=16 observations is drawn from a normal population with µ=1000 and σ=200. Find the following. i) P( >1050) ii) P(960< <1050) Part b) (3 marks) An automatic machine in a manufacturing process is operating properly if the lengths of an important subcomponent are normally distributed with mean=117cm and standard deviation =5.2 cm. i) Find the probability that one selected subcomponent is longer than 120cm. ii) Find the probability that if four subcomponents are randomly selected, their mean length exceeds 120cm. Question 2 [5 marks] Part a) (2 marks) ...

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A sample of n=16 observations is drawn from a normal population with µ=1000 and σ=200. Find the following. i) P( >1050) ii) P(960< <1050) Part b) (3 marks) An automatic machine in a manufacturing process is operating properly if the lengths of an important subcomponent are normally distributed with mean=117cm and standard deviation =5.2 cm. i) Find the probability that one selected subcomponent is longer than 120cm. ii) Find the probability that if four subcomponents are randomly selected, their mean length exceeds 120cm. Question 2 [5 marks] Part a) (2 marks) Calculate the statistic, set up the rejection region, draw the sampling distribution and interpret the result, H0: µ=50 H1: µ<50 Given that: σ=15, n=100, =48, α=0.05. Part b) (3 marks) A business student claims that, on average, an MBA student is required to prepare more than five cases per week. To examine the claim a professor asks a random sample 10 MBA student to report the number of cases they prepare weekly; the professor calculates the mean value and standard deviation, which is 6 and 1.5, respectively. Can the professor conclude at the 5% significance level that the claim is true, assuming that the number of case is normal distribution? Question 3 [4 marks] Part a) (2 marks) Find the following probabilities by checking the z table i) P((Z>-0.8) ii) P(-0.85View Less >>

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Answer 1

a) I) P(X>1050) = 1050-1000/200/sqrt(16) )
= 50/200/4 = 50/50 = 1

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