Question

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.

Answer 1

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%