Answer:
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%.