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 (can be understood as population mean and population standard deviation)

. What is the probability that a randomly selected orange will contain at least 4.8 ounces? .(4DP)

69.7% of oranges will contain at least how many ounces of juice? (2DP)

67% of oranges will contain an amount that is between what two values (symmetrically distributed around the mean)? XL= and XU= . (2DP)

Explanation / Answer

Solution:

We are given that the random variable X follows normal distribution.

Mean = 5

SD = 0.5

Question

What is the probability that a randomly selected orange will contain at least 4.8 ounces? .(4DP)

We have to find P(X4.8)

P(X4.8) = P(X>4.8) = 1 – P(X<4.8)

Z = (X – mean) / SD

Z = (4.8 – 5) / 0.5 = -0.4

P(Z<-0.4) = P(X<4.8) = 0.344578 (by using normal z-table or excel)

P(X4.8) = P(X>4.8) = 1 – P(X<4.8) = 1 - 0.344578 = 0.655422

Required probability = 0.655422

Question

69.7% of oranges will contain at least how many ounces of juice? (2DP)

X = Mean + Z*SD

For upper 69.7% probability, Z = -0.515792

(by using normal z-table or excel command =normsinv(1 – 0.697))

X = 5 + (-0.51579)*0.5

X = 5 – 0.2579 = 4.742104

69.7% of oranges will contain at least 4.74 ounces of juice.

Question

67% of oranges will contain an amount that is between what two values (symmetrically distributed around the mean)? XL= and XU= . (2DP)

Total area = 1

Middle area = 67% = 0.67

Remaining area = 1 – 0.67 = 0.33

Area at left side = 0.33/2 = 0.165

Area at right side = 0.33/2 = 0.165

Z value at left side = -0.97411 (using z-table or excel command =normsinv(0.165))

Z value at right side = 0.974114 (using z-table or excel command =normsinv(0.165+0.67))

X = Mean + Z*SD

XL = 5 -0.97411*0.5 = 4.512945

XU= 5 + 0.97411*0.5 =5.487055

67% of oranges will contain an amount that is between 4.51 and 5.49 ounces approximately.

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