Let $Z \sim N(0,1)$. Find $P(-1.96 < Z < 1.96)$.
Answer (2)
The probability that a standard normal variable Z is between -1.96 and 1.96 is approximately 0.95. This indicates that about 95% of the values in a standard normal distribution fall within this range. This result is commonly used in statistical analysis and hypothesis testing.
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Express the desired probability using the CDF: P ( − 1.9 < Z < − 1.8 ) = Φ ( − 1.8 ) − Φ ( − 1.9 ) .
Find the CDF values: Φ ( − 1.8 ) ≈ 0.03593 and Φ ( − 1.9 ) ≈ 0.02872 .
Calculate the difference: 0.03593 − 0.02872 = 0.00721 .
The probability is: 0.00721 .
Explanation
Understand the problem and provided data We are given that Z follows a standard normal distribution, denoted as Z ∼ N ( 0 , 1 ) . This means Z has a mean of 0 and a standard deviation of 1. We want to find the probability that Z lies between − 1.9 and − 1.8 , which is written as P ( − 1.9 < Z < − 1.8 ) .
Express the probability using CDF To find P ( − 1.9 < Z < − 1.8 ) , we need to calculate the area under the standard normal curve between these two values. We can express this probability as the difference between the cumulative distribution function (CDF) values at − 1.8 and − 1.9 . The CDF, denoted by Φ ( z ) , gives the probability that Z is less than or equal to z . Therefore, we have: P ( − 1.9 < Z < − 1.8 ) = Φ ( − 1.8 ) − Φ ( − 1.9 ) where Φ ( z ) is the CDF of the standard normal distribution.
Find the CDF values Now we need to find the values of Φ ( − 1.8 ) and Φ ( − 1.9 ) . These values can be found using a standard normal table or a calculator with statistical functions. Using a calculator, we find: Φ ( − 1.8 ) ≈ 0.03593 and Φ ( − 1.9 ) ≈ 0.02872
Calculate the probability Now we subtract the CDF values to find the desired probability: P ( − 1.9 < Z < − 1.8 ) = Φ ( − 1.8 ) − Φ ( − 1.9 ) ≈ 0.03593 − 0.02872 = 0.00721
State the final answer Therefore, the probability that Z lies between − 1.9 and − 1.8 is approximately 0.00721 .
Examples
Understanding probabilities related to the standard normal distribution is crucial in many real-world applications. 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 bolts with a mean diameter of 10mm and a standard deviation of 0.1mm, you can use the standard normal distribution to calculate the probability that a randomly selected bolt has a diameter between 9.8mm and 9.9mm. This helps ensure product quality and consistency.
