Assume the weight of Valencia oranges is normally distributed with a mean 9 oz and standard deviation 2 oz. What is the probability that a sample of 100 units show a mean weight of less than 9.5 oz?

0.99379

Step-by-step explanation:

The first thing to do here is to calculate the z-score

mathematically;

z-score = x-mean/SD/√(n)

From the question x = 9.5 ,

mean = 9, SD = 2 and n = 100

Plugging the values we have;

z-score = (9.5-9)/2/√(100) = 0.5/2/10 = 0.5/0.2 = 2.5

So the probability we want to calculate is;

P(z<2.5)

We use the standard table for this

and that equals 0.99379

y=mx+b

step-by-step explanation:

m slope

b y-intercept

step-by-step explanation:

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