33. A factory makes low cost soccer balls with diameters that are normally distributed with mean \( \mu \mathrm{cm} \) and standard deviation 0.25 cm . (a) Estimate the number of balls in a random sample of 200 balls that have diameters within 0.3 cm of the mean diameter. (b) The diameters of \( 80 \% \) of the soccer balls produced by this factory have diameters within \( k \mathrm{~cm} \) of the mean diameter. Find the value of \( k \). (c) \( 25 \% \) of the balls have diameters less than 21.832 cm . Find \( \mu \).

To solve the problem, we will use properties of the normal distribution. ### (a) Estimate the number of balls in a random sample of 200 balls that have diameters within 0.3 cm of the mean diameter. Given: - Mean diameter \( \mu \) - Standard deviation \( \sigma = 0.25 \) cm - We want to find the proportion of balls within \( \mu \pm 0.3 \) cm. First, we calculate the z-scores for the limits \( \mu - 0.3 \) and \( \mu + 0.3 \): \[ z_1 = \frac{(\mu - 0.3) - \mu}{0.25} = \frac{-0.3}{0.25} = -1.2 \] \[ z_2 = \frac{(\mu + 0.3) - \mu}{0.25} = \frac{0.3}{0.25} = 1.2 \] Next, we look up the z-scores in the standard normal distribution table or use a calculator to find the probabilities: - For \( z_1 = -1.2 \), the cumulative probability \( P(Z < -1.2) \approx 0.1151 \). - For \( z_2 = 1.2 \), the cumulative probability \( P(Z < 1.2) \approx 0.8849 \). Now, we find the probability of a ball being within \( \mu \pm 0.3 \): \[ P(\mu - 0.3 < X < \mu + 0.3) = P(Z < 1.2) - P(Z < -1.2) \approx 0.8849 - 0.1151 = 0.7698 \] Now, we estimate the number of balls in a sample of 200: \[ \text{Number of balls} = 200 \times 0.7698 \approx 153.96 \approx 154 \] ### (b) The diameters of \( 80\% \) of the soccer balls produced by this factory have diameters within \( k \) cm of the mean diameter. Find the value of \( k \). To find \( k \), we need to find the z-score that corresponds to the cumulative probability of \( 0.90 \) (since \( 80\% \) of the balls are within \( k \) cm, we want the middle \( 80\% \), which leaves \( 10\% \) in each tail). From the z-table, the z-score for \( 0.90 \) is approximately \( 1.28 \). Now we can calculate \( k \): \[ k = z \cdot \sigma = 1.28 \cdot 0.25 = 0.32 \text{ cm} \] ### (c) \( 25\% \) of the balls have diameters less than 21.832 cm. Find \( \mu \). We need to find the z-score that corresponds to the cumulative probability of \( 0.25 \). From the z-table, the z-score for \( 0.25 \) is approximately \( -0.674 \). Using the z-score formula: \[ z = \frac{X - \mu}{\sigma} \] Substituting the known values: \[ -0.674 = \frac{21.832 - \mu}{0.25} \] Now, we can solve for \( \mu \): \[ 21.832 - \mu = -0.674 \cdot 0.25 \] \[ 21.832 - \mu = -0.1685 \] \[ \mu = 21.832 + 0.1685 = 22.0005 \text{ cm} \] ### Summary of Results: (a) Approximately 154 balls have diameters within 0.3 cm of the mean. (b) The value of \( k \) is 0.32 cm. (c) The mean diameter \( \mu \) is approximately 22.0005 cm.