Question 1 of 10
What is the probability that a data value in a normal
distribution is between a zscore of -1.52 and a z-score of -
0.34? Round your answer to the nearest tenth of a percent.
A. 28.3%
B. 30.3%
C. 29.3%
D. 27.3%
SUBMIT

Answer:30.3

Step-by-step explanation:

30.3% is the probability that a data value in a normal distribution is between a z-score of -1.52 and a z-score of -0.34.

Probability is a number that expresses the likelihood or chance that a specific event will take place. Both proportions ranging from 0 to 1 and percentages ranging from 0% to 100% can be used to describe probabilities.

We can use a standard normal distribution table or calculator to find the probability that a data value falls between two given z-scores.

The probability of a data value being between a z-score of -1.52 and a z-score of -0.34 is equal to the area under the standard normal distribution curve between these two z-scores. We can find this area by subtracting the area to the left of -0.34 from the area to the left of -1.52:

P(-1.52 < z < -0.34) = P(z < -0.34) - P(z < -1.52)

Using a standard normal distribution table or calculator, we can find that P(z < -0.34) is approximately 0.3665 and P(z < -1.52) is approximately 0.0643. Therefore, the probability of a data value being between a z-score of -1.52 and a z-score of -0.34 is:

P(-1.52 < z < -0.34) = 0.3665 - 0.0643 = 0.3022

Rounding this answer to the nearest tenth of a percent, we get:

P(-1.52 < z < -0.34) ≈ 30.2%

Therefore, the closest answer choice is B, 30.3%.

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