A machine used to extract juice from oranges obtains an amount from each orange

Question

A machine used to extract juice from oranges obtains an amount from each orange that is approximately normally distributed, with a mean of 5 ounces and a standard deviation of 0.5 ounces. Suppose that a sample of 30 oranges is selected a What is the probability that the sample mean will be at least 4.8 ounces? b. There is a 69.7 percent chance that the sample average will fall between what two values symmetrically distributed around the population (4DP) mean? and (5DP) c. There is a 67 percent chance that the sample average will be above which value? (5DP)

Explanation / Answer

We can use the Z/normal distribution to solve the problem since a sampl eof 30 oranges is selected

a.

P(X>=4.8)

= P(Z>= (4.8-5)/(.5/sqrt(30))

= P( Z>= -2.19)

= .98574

b. P(-c<=Z<=c) = .697

So, P(Z<=c) = .697/2 +.5 = .8485

So, the (c-5)/.5 = 1.03 ( the Z value for .8485 is 1.03)

Upper bound = c = 1.03*.5/sqrt(30) + 5 = 5.09403

and lower bound = -1.03*.5/sqrt(30) +5 = 4.90597

c.

P(X>=c) = .67,

Therefore, 1-P(X<c) = 1-.67 = .33

So, (c-5)/(.5/sqrt(30) = -.44, (Z for .33 cumulative area is Z=-.44)

So , c = 5-.44*.5/sqrt(30) = 4.95983

So, 67% of value are more than 4.95983

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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