When Lin drives a golf ball, it travels a mean distance of 200 yards with a standard deviation of 20 yards, and the distribution of lengths is approximately normal. Approximately what percentage of her shots will travel less than 180 yards?

The percentage of her shots that will travel less than 180 yards is approximately 15.9%

Step-by-step explanation:

First it will be required to find the standard or z-score from the following formula

Where:

z = z or standard score

x = value that is observed = 180 yards

μ = Sample mean = 200 yards

σ = Sample standard deviation yards

Plugging the values, we have;

Therefore, the probability that her shots will travel less than 180 is given by the relation;

p( x < 180 yards) = p(z < -1) = 0.15866

Therefore, the percentage of her shots that will travel less than 180 = 0.15866 × 100

= 15.866% ≈ 15.9%.