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Find the​ z-score such that the interval within z standard deviations of the mean for a normal distribution contains a. 30​% of the probability. b. 77​% of the probability.

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Final answer:

The question asks for the z-scores corresponding to intervals containing 30% and 77% of the probability in a normal distribution. Using a z-table, one can find these specific z-scores that capture the desired central probabilities, with a reminder that the empirical rule provides approximate values for different standard deviations.

Step-by-step explanation:

The question involves finding the z-scores for particular probabilities within a normal distribution. To begin, we'll tackle part a) which asks for the z-score corresponding to a probability of 30%. Since most z-tables show the area to the left of the z-score, you would look for an area of 0.15 (which, when doubled, gives the central 30%) and find the corresponding z-score.

For part b), which wants the z-score that corresponds to 77% of the probability, we determine the area in each tail by subtracting 77% from 100% and then dividing by 2, giving us 11.5% in each tail. We then find the area to the left of the z-score, which would be 100% - 11.5% = 88.5%, corresponding to the z-score.

As a rule of thumb, known as the empirical rule or the 68-95-99.7 rule, we know about 68% of values fall within z-scores of -1 and +1, about 95% within -2 and +2, and about 99.7% within -3 and +3. However, for our specific percentages of 30% and 77%, we would need to consult the z-table directly to find the exact z-scores.

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