Final answer:
The probability P(x<73.3) for a normally distributed variable with a z-score of 0.83 is approximately 0.7967 when rounded to four decimal places.
Step-by-step explanation:
If x is normally distributed and the z-score of the x-value 73.3 is 0.83, then P(x<73.3) represents the probability that a random variable x is less than 73.3. To find P(x<73.3), we use the z-score and the properties of the standard normal distribution. A z-score of 0.83 corresponds to a certain area to the left under the standard normal curve.
Referencing a z-table or using statistical software, we find that the area to the left of a z-score of 0.83 is typically around 0.7967. Since the area to the left under the curve represents the probability P(x<73.3), we can write P(x<73.3) = 0.7967. This value should be rounded to four decimal places as per the question's requirement.
Hence, the probability that x is less than 73.3, given that x is normally distributed with a z-score of 0.83 for the value 73.3, is approximately 0.7967 when rounded to four decimal places.