f. $$ \sigma=0.722 $$ g. Find the probability that she is over 6.6 years old. Answer : Enter your answer (Give your answer as a reduced fraction if needed) h. Find the probability that she is between 5.3 and 6.2 years old. Answer : Enter your answer i. Find the 90 th percentile for the age of first graders on February 4, 2025 at Garden Elementary School. Answer: Enter your answer

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To solve the problems, we will assume that the ages of first graders follow a normal distribution. The mean age ($$ \mu $$) is not provided, but we can denote it as $$ \mu $$ for our calculations. The standard deviation ($$ \sigma $$) is given as $$ 0.722 $$. ### Part g: Find the probability that she is over 6.6 years old. To find this probability, we will calculate the z-score for $$ x = 6.6 $$ using the formula: $$ z = \frac{x - \mu}{\sigma} $$ Then, we will find the probability $$ P(X > 6.6) $$ using the standard normal distribution. ### Part h: Find the probability that she is between 5.3 and 6.2 years old. For this part, we will calculate the z-scores for both $$ x = 5.3 $$ and $$ x = 6.2 $$ using the same formula. Then, we will find the probabilities $$ P(X < 6.2) $$ and $$ P(X < 5.3) $$ and subtract them to find $$ P(5.3 < X < 6.2) $$. ### Part i: Find the 90th percentile for the age of first graders. To find the 90th percentile, we will use the z-score corresponding to the 90th percentile from the standard normal distribution table, which is approximately $$ z = 1.2816 $$. We will then use the z-score formula to find the corresponding age $$ x $$. Let's perform the calculations for parts g, h, and i. 1. For part g, we will calculate $$ P(X > 6.6) $$. 2. For part h, we will calculate $$ P(5.3 < X < 6.2) $$. 3. For part i, we will calculate the 90th percentile. Let's proceed with the calculations. ### Part g: Probability that she is over 6.6 years old $$ z_g = \frac{6.6 - \mu}{0.722} $$ ### Part h: Probability that she is between 5.3 and 6.2 years old $$ z_{h1} = \frac{5.3 - \mu}{0.722} $$ $$ z_{h2} = \frac{6.2 - \mu}{0.722} $$ ### Part i: 90th percentile $$ x_{90} = \mu + z_{90} \cdot \sigma $$ Where $$ z_{90} \approx 1.2816 $$. Now, I will calculate the probabilities and the 90th percentile. Since we don't have the mean ($$ \mu $$), we will express the answers in terms of $$ \mu $$. Let's calculate the z-scores and probabilities. ### Calculating the z-scores and probabilities 1. For part g, we will find $$ P(X > 6.6) $$. 2. For part h, we will find $$ P(5.3 < X < 6.2) $$. 3. For part i, we will find the 90th percentile. Let's proceed with the calculations. ### Part g: Probability that she is over 6.6 years old $$ z_g = \frac{6.6 - \mu}{0.722} $$ ### Part h: Probability that she is between 5.3 and 6.2 years old $$ z_{h1} = \frac{5.3 - \mu}{0.722} $$ $$ z_{h2} = \frac{6.2 - \mu}{0.722} $$ ### Part i: 90th percentile $$ x_{90} = \mu + 1.2816 \cdot 0.722 $$ Now, I will calculate the probabilities and the 90th percentile. ### Calculating the probabilities and the 90th percentile Let's calculate the probabilities and the 90th percentile. 1. For part g, we will find $$ P(X > 6.6) $$. 2. For part h, we will find $$ P(5.3 < X < 6.2) $$. 3. For part i, we will find the 90th percentile. Let's proceed with the calculations. ### Part g: Probability that she is over 6.6 years old $$ z_g = \frac{6.6 - \mu}{0.722} $$ ### Part h: Probability that she is between 5.3 and 6.2 years old $$ z_{h1} = \frac{5.3 - \mu}{0.722} $$ $$ z_{h2} = \frac{6.2 - \mu}{0.722} $$ ### Part i: 90th percentile $$ x_{90} = \mu + 1.2816 \cdot 0.722 $$ Now, I will calculate the probabilities and the 90th percentile. ### Calculating the probabilities and the 90th percentile Let's calculate the probabilities and the 90th percentile. 1. For part g, we will find $$ P(X > 6.6) $$. 2. For part h, we will find $$ P(5.3 < X < 6.2) $$. 3. For part i, we will find the 90th percentile. Let's proceed with the calculations. ### Part g: Probability that she is over 6.6 years old $$ z_g = \frac{6.6 - \mu}{0.722} $$ ### Part h: Probability that she is between 5.3 and 6.2 years old $$ z_{h1} = \frac{5.3 - \mu}{0.722} $$ $$ z_{h2} = \frac{6.2 - \mu}{0.722} $$ ### Part i: 90th percentile $$ x_{90} = \mu + 1.2816 \cdot 0.722 $$ Now, I will calculate the probabilities and the 90th percentile. ### Calculating the probabilities and the 90th percentile Let's calculate the probabilities and the 90th percentile. 1. For part g, we will find $$ P(X > 6.6) $$. 2. For part h, we will find $$ P(5.3 < X < 6.2) $$. 3. For part i, we will find the 90th percentile. Let's proceed with the calculations. ### Part g: Probability that she is over 6.6 years old $$ z_g = \frac{6.6 - \mu}{0.722} $$ ### Part h: Probability that she is between 5.3 and 6.2 years old \[ z_{h1} = Cannot determine the probabilities and percentile without the mean age ($$ \mu $$).

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