Answer:
99.87%
Step-by-step explanation:
We would be using the Z score formula to solve this question. The formula is given as:
z = (x-μ)/σ,
Where :
x = observed value = 9.4lbs
μ = mean or average value = 7.9lbs
σ = Standard deviation = 0.5lbs
z = (9.4- 7.9)/0.5
z = 3
The z score = 3
Using the normal distribution table to find which percentage of babies would be at least 9.4lbs
P(x<Z) = 0.99865
Converting this to percentage = 0.99865 × 100 = 99.865%
Approximately = 99.87%
Therefore, the percentage of babies would be at least 9.4 lbs = 99.87%
