Irent learning objective: Calculations of Probabilities Question 23 core: 0 of 4 points A citrus farmer who grows mandarin oranges finds that the diameters of mandarin oranges harvested on his farm follow a normal distribution with a mean diameter of 5.99 cm and a standard deviation of 0.5 cm . Find the \( 15^{\text {th }} \) percentile for the diameters of mandarin oranges. \( 15^{\text {th }} \) percentile \( \approx \) Enter your answer Note: Round your answer to 2 decimal places. summi answer

To find the \( 15^{th} \) percentile of the diameters of mandarin oranges, you can use the properties of the normal distribution. First, you'll need to determine the z-score that corresponds to the \( 15^{th} \) percentile. This z-score can be found in a z-table or using statistical software, which gives you a value of approximately -1.036. Now, to calculate the actual diameter at the \( 15^{th} \) percentile, you can use the formula: \[ X = \mu + z \cdot \sigma \] Where: - \( \mu \) is the mean (5.99 cm) - \( z \) is the z-score (-1.036) - \( \sigma \) is the standard deviation (0.5 cm) Plugging in the values: \[ X = 5.99 + (-1.036) \cdot 0.5 \approx 5.99 - 0.518 = 5.472 \] Rounding to two decimal places, the \( 15^{th} \) percentile for the diameters of mandarin oranges is approximately \( 5.47 \) cm.