Averages of several measurements are less variable than individual measurements.

Question

Averages of several measurements are less variable than individual measurements. The true mass of the whiskey sample is 4550 grams, or 4.55 milligrams (mg). Antonio's measurements have the normal distribution with mean 4.55 mg and standard deviation 6.8 mg. In this case, the mean of his 6 measurements also has a normal distribution.

What is the probability that Antonio misses the true mass by more than 6.46 mg in either direction if he makes one measurement?

Use 4 decimal places.

Response:


d.(1 pts)
What is the probability that the mean of 6 independent measurements misses the true mass by more than 6.46 mg?

Use 4 decimal places.

Response:

Explanation / Answer

a)
µ = 4.55
s = 6.8
standardize x to z = (x - µ) / s
P(x > 11.01) = P( z > (11.01-4.55) / 6.8)
= P(z > 0.95) = 0.1711
(From Normal probability table)

P( z < -0.95) = 0.1711
P( miss in either direction) = 0.1711(2) = 0.3422

b)
Mean µ = 4.55
Standard deviation s = 6.8
Standard error s / sqrt n = 6.8 / sqrt 6 = 2.7760884
standardize xbar to z = (xbar - µ) / (s / sqrt n )
P(xbar > 11.01) = P( z > (11.01-4.55) / 2.776088)
= P(z > 2.327) = 0.0099
(from normal probability table)

P( z < -2.327) = 0.0099
P(missing ) = (0.0099)(2) = 0.0198

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.