Answer:
(4) 111
Step-by-step explanation:
Given that:
mean (μ) = 6.4 pounds
Standard deviation (σ) = 0.8 pounds
Number of Siamese cats (n) = 128
The z score (z) is given by the equation:
[tex]z=\frac{x-\mu}{\sigma}[/tex], Where x is the raw score.
For a raw score of 5.2 pounds:
[tex]z=\frac{x-\mu}{\sigma}=\frac{5.2-6.4}{0.8}=-1.5[/tex]
For a raw score of 7.6 pounds:
[tex]z=\frac{x-\mu}{\sigma}=\frac{7.6-6.4}{0.8}=1.5[/tex]
From the normal probability table:
P(5.2 < x < 7.6) = P(-1.5 < z < 1.5) = P(z < 1.5) - P(z < -1.5) = 0.9332 - 0.0668= 0.8664 = 86.64%
Therefore 86.64% of Siamese cats have weights between 5.2 and 7.6 pounds.
The number of Siamese cats that have weights between 5.2 and 7.6 pounds = 86.64% × n = 0.8664 × 128 ≈ 111
