If the hourly wage of detasselers is normally distributed with a mean of $\backslash (13$ and a standard deviation of $\backslash )2$, what is the probability that a randomly selected detasseler makes less than $\$13$ per hour? Express your answer as $p(z<0.5)$.

• A. $0.8413$
• B. $0.6915$
• C. $0.5$
• D. $0.3085$

Answer: (C)

Verified step by step guidance

1. Identify the given parameters: the mean hourly wage \( \mu = 13 \) dollars and the standard deviation \( \sigma = 2 \) dollars. The distribution of wages is normal.
2. Recognize that the problem asks for the probability that a randomly selected detasseler makes less than \$13 per hour, which means finding \( P(X < 13) \) where \( X \) is the wage.
3. Standardize the value \( X = 13 \) to a \( z \)-score using the formula: \[ z = \frac{X - \mu}{\sigma} \] Substitute the values to get the \( z \)-score corresponding to \$13.
4. Since \( X = 13 \) is exactly the mean, the \( z \)-score will be zero. The probability \( P(X < 13) \) is then the same as \( P(Z < 0) \) for the standard normal distribution.
5. Recall that the standard normal distribution is symmetric about zero, so \( P(Z < 0) = 0.5 \). This means there is a 50% chance a randomly selected detasseler earns less than \$13 per hour.