Let Z be a standard normal random variable with mean 0 and variance 1. What is the probability that Z lies between 0 and 3.01, that is, 0 less than Z less than 3.01?

Introduction / Context:
This question involves the standard normal distribution, which is a continuous probability distribution with mean 0 and variance 1. The variable Z denotes a standard normal random variable. We are asked to find the probability that Z takes a value between 0 and 3.01. Such problems require the use of standard normal probability tables or a calculator that can evaluate cumulative distribution values.

Given Data / Assumptions:

• Z is a standard normal random variable.
• Mean of Z is 0 and standard deviation is 1.
• We are interested in P(0 less than Z less than 3.01).
• Standard normal tables typically give cumulative probabilities from negative infinity up to a given Z value.
• The standard normal distribution is symmetric around zero.

Concept / Approach:
To find P(0 less than Z less than 3.01), we can use the cumulative distribution function values. The expression equals F(3.01) minus F(0), where F denotes the standard normal cumulative distribution function. Because of symmetry, F(0) equals 0.5. The value F(3.01) is very close to 1 and is found from a standard table or calculator. Subtracting 0.5 from that cumulative value gives the required probability between 0 and 3.01.

Step-by-Step Solution:
Step 1: Express the desired probability as P(0 less than Z less than 3.01) = F(3.01) minus F(0).Step 2: Recall that for the standard normal distribution, F(0) = 0.5 because the distribution is symmetric around zero.Step 3: Use a standard normal table or calculator to find F(3.01). The value is approximately 0.9987.Step 4: Substitute these values into the expression to get P(0 less than Z less than 3.01) = 0.9987 minus 0.5.Step 5: Perform the subtraction: 0.9987 minus 0.5 = 0.4987.Step 6: Therefore, the probability that Z lies between 0 and 3.01 is approximately 0.4987.
Verification / Alternative check:
As a rough check, we know that almost all of the standard normal distribution lies within three standard deviations of the mean. The probability from minus 3 to 3 is about 0.9973, so the probability from 0 to 3 should be roughly half of that, about 0.4987. Extending 3 slightly to 3.01 hardly changes the value. This back of the envelope reasoning supports the more precise calculation that uses F(3.01) and gives confidence in the result.

Why Other Options Are Wrong:
0.5 would be the probability from 0 to infinity, not from 0 to 3.01. 0.9987 is the cumulative probability from negative infinity up to 3.01, not between 0 and that point. 0.1217 and 0.0013 are far too small and do not match any realistic central probability range for values between 0 and a large positive Z value like 3.01.

Common Pitfalls:
Students sometimes forget to subtract F(0) and instead read the cumulative probability directly from the table, which overestimates the required value. Another common error is to confuse the probability from 0 to 3 with the probability from minus 3 to 3 and not take advantage of symmetry. Carefully translating the interval into differences of cumulative probabilities is essential.

Final Answer:
The probability that the standard normal variable Z lies between 0 and 3.01 is approximately 0.4987.