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We solve each part using the binomial distribution formula and approximations where necessary.
a) less than 32?
P(X < 32) = \sum_{k=0}^{31} \binom{400}{k} (0.1)^k (0.9)^{400-k}
Using normal approximation:
\mu = np = 400 \times 0.1 = 40
\sigma = \sqrt{np(1-p)} = \sqrt{400 \times 0.1 \times 0.9} = \sqrt{36} = 6
Normal approximation to the binomial:
P(X < 32) \approx P\left(\frac{X - 40}{6} < \frac{32 - 40}{6}\right) = P\left(Z < -\frac{8}{6}\right) = P\left(Z < -1.33\right) \approx 0.0918
So:
P(X < 32) \approx 0.0918
b) greater than 49?
P(X > 49) = 1 - P(X \leq 49)
Using normal approximation:
P(X \leq 49) \approx P\left(\frac{X - 40}{6} \leq \frac{49 - 40}{6}\right) = P\left(Z \leq 1.5\right) \approx 0.9332
So:
P(X > 49) \approx 1 - 0.9332 = 0.0668
e) at least 35 but less than 47?
P(35 \leq X < 47) = P(X < 47) - P(X < 35)
Using normal approximation:
P(X < 47) \approx P\left(\frac{X - 40}{6} < \frac{47 - 40}{6}\right) = P\left(Z < 1.17\right) \approx 0.8790
P(X < 35) \approx P\left(\frac{X - 40}{6} < \frac{35 - 40}{6}\right) = P\left(Z < -0.83\right) \approx 0.2033
So:
P(35 \leq X < 47) = P(X < 47) - P(X < 35) \approx 0.8790 - 0.2033 = 0.6757
Therefore:
a) P(X < 32) \approx 0.0918
b) P(X > 49) \approx 0.0668
e) P(35 \leq X < 47) \approx 0.6757