the average newborn baby weighs 7.9 lbs with standard deviation of approximately 0.5lbs. if this is normal distribution then what percentage of babies would be at least 9.4 lbs.

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%

h who knows

step-by-step explanation: