Final answer:
The probability that a normal random variable with a mean of 6 and a variance of 4 lies between 4 and 7 is found using z-scores and is approximately 0.5328.
Step-by-step explanation:
The student's question asks about the probability that a normal random variable with a mean of 6 and a variance of 4 lies between the values 4 and 7. To find this, we first need to calculate the standard deviation, which is the square root of the variance. The standard deviation in this case is 2 (since 2 squared equals 4). Then, we can use z-scores to find the probability by standardizing the values of 4 and 7:
- For x = 4, z = (4 - 6) / 2 = -1.
- For x = 7, z = (7 - 6) / 2 = 0.5.
Next, we look up these z-scores in the standard normal distribution table (or use technology like a calculator or software for the exact values) to find the probabilities associated with each z-score. The probability that x is less than or equal to 4 is the same as the probability of a z-score less than or equal to -1, and the probability that x is less than or equal to 7 is the same as the probability of a z-score less than or equal to 0.5.
The probability that x lies between 4 and 7 is the difference between these two probabilities. For a z-score of -1, the probability is approximately 0.1587 and for a z-score of 0.5, it is approximately 0.6915. Subtracting these gives:
P(4 < x < 7) = P(z < 0.5) - P(z < -1)
= 0.6915 - 0.1587
= 0.5328.
Therefore, the probability that x lies between 4 and 7 is closest to 0.5328.