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

the cat is blue because the probability of extension towards the prefix of the diameter of the circle could only mathematically equal to the sum of the roots of the carrots from   your garden.

step-by-step explanation: