Question: Name______________________________ Date________________ 1. Find the Z-Score for a $\mu = 34$, $\sigma = 5.1$, $x = 17$ 2. Find the Z-Score for a $\mu = 1055$, $\sigma = 211$, $x = 1150$ 3. The mean number of seeds for a population of sunflowers is $1400$, and the standard deviation is $300$. What percentage of flowers will have less than $1500$ seeds? 4. What percentage of sunflowers from the same population have more than $1000$ seeds? 5. What percentage of sunflowers from the same population would have between $1000$ and $1700$ seeds?

1. **Find the Z-Score for $\mu = 34$, $\sigma = 5.1$, $x = 17$** The formula for the Z-Score is: $$Z = \frac{x - \mu}{\sigma}$$ Step 1: Substitute the values: $$Z = \frac{17 - 34}{5.1}$$ Step 2: Calculate the numerator: $$17 - 34 = -17$$ Step 3: Write the fraction: $$Z = \frac{-17}{5.1}$$ Step 4: Simplify the fraction: $$Z = -\frac{17}{5.1}$$ Step 5: Calculate the decimal value: $$Z \approx -3.33$$ --- 2. **Find the Z-Score for $\mu = 1055$, $\sigma = 211$, $x = 1150$** Step 1: Use the formula: $$Z = \frac{x - \mu}{\sigma}$$ Step 2: Substitute values: $$Z = \frac{1150 - 1055}{211}$$ Step 3: Calculate numerator: $$1150 - 1055 = 95$$ Step 4: Write the fraction: $$Z = \frac{95}{211}$$ Step 5: Calculate decimal value: $$Z \approx 0.45$$ --- 3. **Percentage of flowers with less than $1500$ seeds** Step 1: Calculate Z-Score for $x=1500$: $$Z = \frac{1500 - 1400}{300} = \frac{100}{300} = \frac{1}{3} \approx 0.33$$ Step 2: Use standard normal distribution table or calculator to find the percentage for $Z=0.33$. Step 3: From the table, $P(Z < 0.33) \approx 0.6293$ or $62.93\%$. So, about $62.93\%$ of flowers have less than $1500$ seeds. --- 4. **Percentage of flowers with more than $1000$ seeds** Step 1: Calculate Z-Score for $x=1000$: $$Z = \frac{1000 - 1400}{300} = \frac{-400}{300} = -\frac{4}{3} \approx -1.33$$ Step 2: Find $P(Z < -1.33)$ from the table, which is about $0.0918$ or $9.18\%$. Step 3: Since we want more than $1000$ seeds, calculate: $$P(Z > -1.33) = 1 - P(Z < -1.33) = 1 - 0.0918 = 0.9082$$ So, about $90.82\%$ of flowers have more than $1000$ seeds. --- 5. **Percentage of flowers with seeds between $1000$ and $1700$** Step 1: Calculate Z-Scores: For $x=1000$: $$Z_1 = \frac{1000 - 1400}{300} = -1.33$$ For $x=1700$: $$Z_2 = \frac{1700 - 1400}{300} = 1$$ Step 2: Find probabilities from the table: $$P(Z < 1) = 0.8413$$ $$P(Z < -1.33) = 0.0918$$ Step 3: Calculate the percentage between: $$P(-1.33 < Z < 1) = P(Z < 1) - P(Z < -1.33) = 0.8413 - 0.0918 = 0.7495$$ So, about $74.95\%$ of flowers have between $1000$ and $1700$ seeds.