7. Worked Examples

Let's delve into how we can determine the probability that a random variable falls between two values using the normal distribution. For instance, we have X distributed as N(50, 8). Who can tell me the first step?

Do we need to standardize the values by converting them to Z-scores?

Exactly! Standardization is key. To standardize, we use the formula Z = (X - μ) / σ. Can anyone calculate the Z-scores for 42 and 58?

For 42, it would be Z = (42 - 50) / 8, which is -1.

And for 58, Z = (58 - 50) / 8, which is 1.

Great! Now we look these values up in the Z-table. What do you find for P(Z < 1) and P(Z < -1)?

I have 0.8413 for P(Z < 1) and 0.1587 for P(Z < -1).

Fantastic! So, to find the probability that X is between 42 and 58, what do we do next?

We subtract the two probabilities: 0.8413 - 0.1587.

Correct! What does that give us?

0.6826 or 68.26%.

Well done! Remember, around 68% of values fall within one standard deviation of the mean, fitting our example nicely.

Now, let's tackle a different problem. We need to find the minimum score that places a student in the top 5% of test scores distributed as N(70, 12). What should our first step be?

We need to find the Z-score for the 95th percentile since top 5% means we want scores above that threshold.

Correct! The Z-score for the 95th percentile is approximately 1.645. Can anybody calculate the actual score using this Z-value?

We can use the formula x = μ + z * σ. So it would be x = 70 + 1.645 * 12.

Excellent calculation! What do you get when you perform that operation?

It would be x = 70 + 19.74, which equals 89.74.

Good! So, for a student to be in the top 5%, they need a minimum score of 89.74. Does this calculation method make sense?

Yes, we understand that we are standardizing the scores and using the Z-table to find our cutoff.

Great job! Remember, understanding how to standardize and find specific probabilities is fundamental in statistics.