Answer:
0.0421
Step-by-step explanation:
We solve using the z score formula
z-score is is z = (x-μ)/σ,
where x is the raw score
μ is the population mean = μ = 8.9 pounds
σ is the population standard deviation = σ = 1.1 pounds
For x < 7 pounds
z = 7 - 8.9/1.1
= -1.72727
Probability value from Z-Table:
P(x<7) = 0.042059
Therefore, the proportion of domestic house cats weigh less than 7 pounds is
0.042059
Approximately = 0.0421