13 The weights of bananas from a certain grower are approximately normally distributed. If the standard deviation of the weight of these bananas is 5 g , and \( 16 \% \) of the bananas weigh less than 96 g , what is the mean weight of the bananas?

To find the mean weight of the bananas, we can use the properties of the normal distribution. Since 16% of the bananas weigh less than 96 g, this means that 96 g is the 16th percentile of the distribution. In a standard normal distribution, the z-score that corresponds to the 16th percentile is approximately -1.00. Using the formula for z-score: \[ z = \frac{(X - \mu)}{\sigma} \] where \( X \) is the value (96 g), \( \mu \) is the mean, and \( \sigma \) is the standard deviation (5 g), we can set up the equation: \[ -1.00 = \frac{(96 - \mu)}{5} \] Solving for \( \mu \): \[ -5 = 96 - \mu \] \[ \mu = 101 \text{ g} \] So, the mean weight of the bananas is 101 g.