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For a standard normal distribution, find: P(z>-0.03). Express the probability as a decimal rounded to 4 decimal places.

User Octoshape
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P(z > -0.03) for a standard normal distribution is approximately 0.5112.

In the context of a standard normal distribution, the value "z" represents the number of standard deviations a data point is from the mean. In this case, we're interested in finding the probability that a randomly selected value is greater than -0.03 standard deviations from the mean.

The standard normal distribution is symmetric around zero, and the probability density function is bell-shaped. Given this symmetry, the area under the curve to the right of -0.03 is equivalent to the area to the left of 0.03. To find this probability, we can consult a standard normal distribution table or use a statistical calculator.

The calculated probability, approximately 0.5112, implies that over half of the distribution lies to the right of -0.03 standard deviations. Therefore, there is about a 51.12% chance that a randomly chosen value from the standard normal distribution is greater than -0.03 standard deviations away from the mean. This probability provides valuable insights into the relative position of a data point within the distribution and is fundamental in statistical analysis and hypothesis testing.

User Fakeer
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