Question 230877: The diameters of grapefruits in a certain orchard are normally distributed with a mean of 6.85 inches and a standard deviation of 0.80 inches. Show all work.
(A) What percentage of the oranges in this orchard have diameters less than 6.3 inches?
(B) What percentage of the oranges in this orchard are larger than 6.75 inches?
(C) A random sample of 100 oranges is gathered and the mean diameter obtained was 6.75. If another sample of 100 is taken, what is the probability that its sample mean will be greater than 6.75 inches?
(D) Why is the z-score used in answering (A), (B), and (C)?
(E) Why is the formula for z used in (C) different from that used in (A) and
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! The diameters of grapefruits in a certain orchard are normally distributed with a mean of 6.85 inches and a standard deviation of 0.80 inches. Show all work.
(A) What percentage of the oranges in this orchard have diameters less than 6.3 inches?
z(6.5)=(6.3-6.85)/0.8 = -0.6875
P(x< 6.3) + P(z < -0.6875) = 24.59%
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(B) What percentage of the oranges in this orchard are larger than 6.75 inches?
Same process. Fid the z-value of 6.75
Find the percentages greater than that z-value.
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(C) A random sample of 100 oranges is gathered and the mean diameter obtained was 6.75. If another sample of 100 is taken, what is the probability that its sample mean will be greater than 6.75 inches?
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The std has changed to 0.8/sqrt(100) = 0.08
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z(6.75) = (6.75-6.85)/0.08 = -1.25
P(x > 6.75) = (z > -1.25) = 0.8944
(D) Why is the z-score used in answering (A), (B), and (C)?
I'll leave that to you. It's probably explained in your text.
(E) Why is the formula for z used in (C) different from that used in (A) and
The sample means have distribution (u,sigma/sqrt(n)) according to
the Central Limit Theorem.
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Cheers,
Stan H.
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