Assume that the weights of ripe watermelons grown at a particular farm are normally distributed with a mean of 30 pounds and a standard deviation of 1.8 pounds. If the farm produces 300 watermelons, how many will weigh less than 30.63 pounds?
Click here to view page 1 of the standard normal distribution table.
Click here to view page 2 of the standard normal distribution table.
Approximately $\square$ watermelons will weigh less than 30.63 pounds.
(Round to the nearest whole number as needed.)
Answer
Final Answer: Approximately \(\boxed{191}\) watermelons will weigh less than 30.63 pounds.
Step 1 ：The problem is asking for the number of watermelons that weigh less than 30.63 pounds. This is a problem of normal distribution. We know that the mean weight of the watermelons is 30 pounds and the standard deviation is 1.8 pounds.
Step 2 ：We can convert the weight of 30.63 pounds to a z-score, which is a measure of how many standard deviations an element is from the mean. The formula for the z-score is \((X - μ) / σ\), where X is the value we are interested in, μ is the mean, and σ is the standard deviation.
Step 3 ：Substituting the given values into the z-score formula, we get \((30.63 - 30) / 1.8 = 0.35\).
Step 4 ：After finding the z-score, we can use a z-table to find the probability that a watermelon weighs less than 30.63 pounds. The probability corresponding to a z-score of 0.35 is approximately 0.6368.
Step 5 ：Finally, we multiply this probability by the total number of watermelons to find the number of watermelons that weigh less than 30.63 pounds. So, \(0.6368 * 300 = 191\).
Step 6 ：Final Answer: Approximately \(\boxed{191}\) watermelons will weigh less than 30.63 pounds.
