Final answer:
The probability that a sunflower will be less than 120 cm tall, given a mean height of 112 cm and standard deviation of 16 cm, is found by first calculating a z-score of 0.5. Looking up this z-score in a standard normal distribution table, we find there is approximately a 69.15% chance that a sunflower will be under 120 cm in height.
Step-by-step explanation:
To find the probability that a sunflower will be less than 120 cm tall, given that the heights are normally distributed with a mean of 112 cm and a standard deviation of 16 cm, we first need to calculate the z-score of the sunflower's height.
The z-score is found through the formula: z = (X - μ) / σ, where X is the value for which we're finding the probability (in this case, 120 cm), μ is the mean, and σ is the standard deviation.
Calculating the z-score:
z = (120 cm - 112 cm) / 16 cm = 0.5
Next, we look up the z-score in a standard normal distribution table or use a calculator that can compute normal probabilities. The table or calculator gives the probability of a z-score being less than 0.5, which approximately corresponds to the sunflower being less than 120 cm tall.
According to standard normal distribution tables, a z-score of 0.5 has a probability of about 0.6915. Therefore, there is a 69.15% chance that a sunflower will be less than 120 cm tall.