The Normal Probability Distribution

Normal Distribution

A telecommunications company has determined that the number of products sold weekly is normally distributed with: - Mean: \(\mu = 900\) - Standard deviation: \(\sigma = 40\)

Let \(X \sim N(900, 40^2)\)

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(i) Estimate the proportion of weeks with more than 960 products sold.

• Compute the z-score: \[ z = \frac{960 - 900}{40} = 1.5 \]
• From standard normal tables: \[ P(X > 960) = P(Z > 1.5) = 1 - P(Z \leq 1.5) = 1 - 0.9332 = \mathbf{0.0668} \]

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(ii) Estimate the proportion of weeks with more than 860 products sold.

• Compute the z-score: \[ z = \frac{860 - 900}{40} = -1.0 \]
• From standard normal tables: \[ P(X > 860) = P(Z > -1.0) = P(Z < 1.0) = 0.8413 \] \[ \mathbf{P(X > 860) = 0.8413} \]

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(iii) Estimate the proportion of weeks with sales between 880 and 960 products.

• Compute z-scores: \[ z_{880} = \frac{880 - 900}{40} = -0.5, \quad z_{960} = \frac{960 - 900}{40} = 1.5 \]
• From tables: \[ P(-0.5 \leq Z \leq 1.5) = P(Z \leq 1.5) - P(Z \leq -0.5) \] \[ = 0.9332 - 0.3085 = \mathbf{0.6247} \]