The length of similar components produced by a company are approximated by a normal distribution model with a mean of 8 cm and a standard deviation of 0.04 cm . Let the random variable X represent the length of one component chosen at random. Without using the normalcdf(...) function, evaluate \( P(X>7.88) \) 99.85 0.003 0.4985 0.997 0.9985

To find \( P(X > 7.88) \), we first standardize the variable using the Z-score formula: \[ Z = \frac{X - \mu}{\sigma} \] Here, \( \mu = 8 \) cm and \( \sigma = 0.04 \) cm. For \( X = 7.88 \): \[ Z = \frac{7.88 - 8}{0.04} = \frac{-0.12}{0.04} = -3 \] Now, we look for \( P(Z > -3) \). Using Z-tables, the area to the left of \( Z = -3 \) is approximately 0.00135. Therefore, \[ P(Z > -3) = 1 - 0.00135 = 0.99865 \] This is approximately \( 0.9985 \). So, the correct answer is: \( 0.9985 \). Now, let's explore some extra fun facts! The concept of a Z-score allows statisticians to compare scores from different normal distributions. It transforms raw scores into a standard scale where we can easily interpret probabilities! Another interesting note is that about 68% of data falls within one standard deviation of the mean in a normal distribution, while about 95% falls within two standard deviations, and a whopping 99.7% falls within three! This properties make normal distribution incredibly useful in various real-world applications such as quality control and risk assessment.