For a standard normal distribution, find:
[tex]$P(-2.56\ \textless \ z\ \textless \ 0.01)$[/tex]

Express the probability as a decimal rounded to 4 decimal places.

The probability that a z-score lies between -2.56 and 0.01 in a standard normal distribution is approximately 0.4988 when rounded to four decimal places.
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Answered by Anonymous | 2025-07-04

Express the desired probability as the difference between CDF values: P ( − 2.56 < z < 0.01 ) = Φ ( 0.01 ) − Φ ( − 2.56 ) .
Find the CDF values: Φ ( 0.01 ) ≈ 0.5040 and Φ ( − 2.56 ) ≈ 0.0052 .
Calculate the probability: 0.5040 − 0.0052 = 0.4988 .
Round the result to 4 decimal places: 0.4988 ​ .

Explanation

Understand the problem We are given a standard normal distribution, and we want to find the probability that the z-score lies between -2.56 and 0.01, which is written as P ( − 2.56 < z < 0.01 ) . This means we need to calculate the area under the standard normal curve between these two z-scores.

Express the probability using CDF To find P ( − 2.56 < z < 0.01 ) , we can use the cumulative distribution function (CDF) of the standard normal distribution, denoted as Φ ( z ) . The CDF gives the probability that a standard normal random variable is less than or equal to z , i.e., P ( Z ≤ z ) = Φ ( z ) .


We can express the desired probability as the difference between the CDF values at the upper and lower bounds:
P ( − 2.56 < z < 0.01 ) = P ( z < 0.01 ) − P ( z < − 2.56 ) = Φ ( 0.01 ) − Φ ( − 2.56 )

Find the CDF values Now, we need to find the values of Φ ( 0.01 ) and Φ ( − 2.56 ) . Using a standard normal distribution table or a calculator, we find:

Φ ( 0.01 ) ≈ 0.503989 and Φ ( − 2.56 ) ≈ 0.005241

Calculate the probability Now we subtract the two CDF values to find the desired probability:

P ( − 2.56 < z < 0.01 ) = 0.503989 − 0.005241 = 0.498748

Round to 4 decimal places Finally, we round the result to 4 decimal places:

P ( − 2.56 < z < 0.01 ) ≈ 0.4987
Examples
Understanding probabilities within a standard normal distribution is crucial in many real-world scenarios. For instance, in quality control, manufacturers use these probabilities to determine the likelihood that a product's measurement falls within acceptable limits. If a machine produces parts with a mean diameter of 5 cm and a standard deviation of 0.1 cm, you can use the standard normal distribution to find the probability that a randomly selected part will have a diameter between 4.8 cm and 5.1 cm. This helps ensure product quality and consistency.

Answered by GinnyAnswer | 2025-07-04