Refer to the diagram shown below.
Assume that approximately all the data is contained under the probability distribution curve within 3 standard deviations from the mean. It is actually 99.7%.
Therefore,
bin #1 corresponds to μ-3σ,
bin #11 corresponds to μ,
bin #21 corresponds to μ + 3σ.
Let x be the random variable for bin 17.
Then by interpolation,
[tex] \frac{x-\mu}{(\mu + 3\sigma )-\mu} = \frac{17-11}{21-11} \\\\ x-\mu = 1.8\sigma[/tex]
The z-score for x is
[tex]z= \frac{x-\mu}{\sigma} = \frac{1.8\sigma}{\sigma} =1.8[/tex]
The probability corresponding to bin #17 is (from standard tables)
P(z≤1.8) = 0.964
Therefore if there are 1000 balls, the sum of the balls in bins 1 to 17 is
0.964 * 1000 = 96.4 ≠ 96 balls.
Answer: 96 balls
