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 4 decimal places, evaluate \( P(X<7.97) \) 0.7734 0.4533 0.5467 0.2266 None of the above. ** Note: Do not round any z-score.

To evaluate \( P(X < 7.97) \), we first need to calculate the z-score using the formula: \[ z = \frac{X - \mu}{\sigma} \] Where: - \( X = 7.97 \) - \( \mu = 8 \) - \( \sigma = 0.04 \) Plugging in the values: \[ z = \frac{7.97 - 8}{0.04} = \frac{-0.03}{0.04} = -0.75 \] Next, we look up the z-score of -0.75 in the standard normal distribution table, or use a calculator, to find \( P(Z < -0.75) \). This gives us approximately 0.2266. So, the answer is **0.2266**. Now for some interesting insights, did you know the concept of the normal distribution was significantly advanced by Carl Friedrich Gauss in the 18th century? His work with the distribution laid the foundation for what we now recognize as the standard normal curve that is ubiquitous in statistics today! Additionally, understanding how to utilize z-scores is a key problem-solving tip in statistics. Many students make the mistake of not knowing how to properly find their corresponding probabilities or confuse the direction of the inequalities when looking up values. Always remember: if you’re looking for \( P(Z < z) \), stick to the cumulative distribution table for your reference!