6. 10 points DevoreSta 5E S02 XP My Notes Ask Your Teacher + The inside diameter

Question

6. 10 points DevoreSta 5E S02 XP My Notes Ask Your Teacher + The inside diameter of a randomly selected piston ring is a random variable with mean value 14 cm and standard deviation 0.04 cm (a) If is the sample mean diameter for a random sample of n = 16 rings, where is the sampling distribution of x centered and what is the standard deviation of the distribution? (Enter your standard deviation to five decimal places.) center standard deviation 64 (b) Answer the questions posed in part (a) for a sample size of n rings. (Enter your standard deviation to ive decimal places.) center standard deviationm (c) For which of the two random samples, the one of part (a) or the one of part (b), is more likely to be within 0.01 cm of 14 cm? Explain your reasoning O is more likely to be within 0.01 cm of 14 cm in sample (a) because of the decreased variability with a smaller sample size. O is more likely to be within 0.01 cm of 14 cm in sample (b) because of the decreased variability with a larger sample size o x is more likely to be within 0.01 cm of 14 cm in sample (b) because of the increased variability with a larger sample size. is more likely to be within 0.01 cm of 14 cm in sample (a) because of the increased variability with a smaller sample size. Need Help? e Tek to Tutr

Explanation / Answer

Solution:-

6) Given that mean = 14cm and sd = 0.04 cm

a) center = 14cm
standard deviation = s/sqrt(n) = 0.04/sqrt(16) = 0.01000

b) center = 14cm
standard deviation = s/sqrt(n) = 0.04/sqrt(64) = 0.00500

c) option D.

7)

a)P(9.99 <= X <= 10.01) = P((9.99-10)/(0.06/sqrt(16) <= Z <= (10.01-10)/(0.06/sqrt(16))
= P(-0.6667 <= Z <= 0.6667)
= 0.4972

b) P(X >= 10.01) = P(Z >= (10.01-10)/(0.06/sqrt(25))

= P ( Z>0.8333 )

=1P ( Z<0.8333 )

=10.7967

=0.2033

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