A juice producer sources all of his oranges from a huge plantation that has a variety of oranges. The amount of juice extracted from each of these oranges has an approximately normal distribution, with a mean of 4.70 ounces and a standard deviation of 0.40 ounces. Suppose you select a random sample of 25 oranges: What is the probability that the sample mean is at least 4.60 ounces?
Question
Answer:
$$\mu=4.70 ounces$$
$$\sigma=0.40 ounces$$
$$n=25$$
$$x=4.60$$
Calculate the z-score:
$$z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}$$
$$z=\frac{4.60-4.70}{\frac{0.40}{\sqrt{25}}}$$
$$z=-\frac{5}{4}$$
Use calculator to calculate its probability:
$$P(x\geq 4.60)=0.89435$$
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5 months ago
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