A turner makes curtain rods that customers custom order. The length of the bars in meters is a random variable T such that T is a normal with expectation 2 and unknown variance. The probability that the length of the bar is greater than 50 cm is 0.9938. A) calculate the variance of T B) calculate the probability that the length of the bar is 2.4 meters

A) To calculate the variance of the normal distribution, we need to use the information given about the probability that the length of the bar is greater than 50 cm. We know that the standard normal distribution (z-distribution) has the property that we can find the probability of a value being greater than a certain point using the z-score. The z-score is calculated as: $$\[ z = \frac{x - \mu}{\sigma} \]$$ Where: - x is the value we're interested in (in this case, 50 cm) - mu is the mean (expectation) of the distribution (given as 2 meters) - $$\( \sigma \)$$ is the standard deviation of the distribution (which is the square root of the variance we're looking for) Since we're given that the probability of the length of the bar being greater than 50 cm is 0.9938, we can find the corresponding z-score using a standard normal distribution table or calculator. The z-score represents how many standard deviations away from the mean the value is. $$\[ P(T > 50) = 0.9938 \]$$ $$\[ z = Z(0.9938) \]$$ (where Z is the z-score) Once we have the z-score, we can use it to find the standard deviation $$\( \sigma \)$$: $$\[ z = \frac{50 \, \text{cm} - 200 \, \text{cm}}{\sigma} \]$$ Now, we can calculate the standard deviation $$\( \sigma \)$$: $$\[ \sigma = \frac{50 \, \text{cm} - 200 \, \text{cm}}{z} \]$$ Finally, the variance $$\( \sigma^2 \)$$is the square of the standard deviation: $$\[ \text{Variance} = \sigma^2 \]$$ B) To calculate the probability that the length of the bar is exactly 2.4 meters, we can use the normal distribution's probability density function (PDF). Since we have the mean mu and variance $$\( \sigma^2 \)$$from part A, we can plug these values into the PDF formula to find the probability. The PDF of a normal distribution is given by: \$$[ f(x) = \frac{1}{\sigma \sqrt{2\pi}} \cdot e^{-\frac{(x - \mu)^2}{2\sigma^2}} \]$$ For this case, x = 2.4 meters. $$\[ P(T = 2.4) = f(2.4) \]$$ Where: - f(x) is the probability density function - x is the value we're interested in (2.4 meters) - mu is the mean (given as 2 meters) - $$\( \sigma \)$$is the standard deviation (calculated in part A) Solving for $$\( P(T = 2.4) \)$$ will give you the probability that the length of the bar is exactly 2.4 meters.