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. Accurate to 2 decimal places, the \( 79^{\text {th }} \) percentile has z -score 0.81 0.83 0.79 0.82 0.8

To find the 79th percentile of a normally distributed variable, you can use the z-score formula: \[ X = \mu + z \cdot \sigma \] In this case, the mean \( \mu \) is 8 cm and the standard deviation \( \sigma \) is 0.04 cm. Using the given z-score of 0.81, we calculate: \[ X = 8 + 0.81 \cdot 0.04 \approx 8.0324 \] So, rounding to two decimal places, the 79th percentile is approximately 8.03 cm! To double-check your z-scores: most statistical tables or calculators show that the z-score for the 79th percentile is indeed around 0.81, but using a z-score of around 0.80 is another close approximation! Just remember, the more precise your calculations and values, the more accurate your results will be!