A box contains 4 pears and 7 oranges. There are 121 similar boxes in a warehouse. One fruit is taken at random from each box.
Using a suitable approximation, find the probability that fewer than 39 are pears.
Solution
Let the random variable \(X\) represent the number of pears. Each box has a probability of \(\frac{4}{11}\) of containing a pear.
The expected number of pears, \(\mu\), is given by:
\(\mu = 121 \times \frac{4}{11} = 44\)
The variance, \(\sigma^2\), is given by:
\(\sigma^2 = 121 \times \frac{4}{11} \times \frac{7}{11} = 28\)
We use a normal approximation to the binomial distribution. The standard deviation \(\sigma\) is:
\(\sigma = \sqrt{28}\)
We apply continuity correction for \(X < 39\), using \(X < 38.5\).
Standardizing, we find:
\(Z = \frac{38.5 - 44}{\sqrt{28}} = -1.039\)
Using the standard normal distribution table, we find:
\(P(Z < -1.039) = \Phi(-1.039) = 1 - 0.8506 = 0.149\)
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