Answer:
21.186%
Step-by-step explanation:
z = (x-μ)/σ,
where
x is the raw score = 2400mg
μ is the population mean = 2000mg
σ is the population standard deviation = 500mg
z = 2400 - 2000/500
z = 0.8
Probability value from Z-Table:
P(x<2400) = 0.78814
P(x>2400) = 1 - P(x<2400) = 0.21186
Converting to percentage:
0.21186 × 100
= 21.186%
Therefore, the percent of the meals
ordered that exceeded the recommended daily allowance of 2400 mg of sodium is 21.186%