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Assume that a randomly selected subject is given a bone density test. Those test scores are normally distributed with a mean of 0 and a standard deviation of 1. Find the probability that a given score is less than 1.99 and draw a sketch of the region.

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Answer:

Step-by-step explanation:

To find this probability, we shall be using the z-score route

Mathematically ;

z-score = (x -mean)/SD

From the question, x = 1.99, mean = 0 and SD = 1

So z = (1.99-0)/1 = 1.99

So the probability we want to calculate is;

P(z<1.99)

This value can be obtained from the standard normal distribution table.

P(z < 1.99) = 0.9767

The sketch of the region is as shown as in the attachment.

Ver imagen Adetunmbiadekunle

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