Question

question 17 - 18: follow steps a - d for each scenario.

1. in a population of sunflowers, the mean number of seeds is 1400, with a standard deviation of 350 seeds. what percentage of sunflowers have less than 1800 seeds? (10 points)

a. give μ and σ: μ:__ σ:__
b. solve for z - score: z = \\(\frac{x - \mu}{\sigma}\\)=
c. find the percent in the calculator:____
d. graph and shade:

1. the age of employees working at a steel mill are normally distributed with a population mean of 33 years old and a standard deviation of 4.5 years. what percent of employees are between 30 and 40 years old? (15 points)

a. give μ and σ: μ:__ σ:__
b. solve for both z - scores: z = \\(\frac{x - \mu}{\sigma}\\)=
z = \\(\frac{x - \mu}{\sigma}\\)=
c. find the percent in the calculator:____
d. graph and shade:

Explanation:

Question 17

Step1: Identify mean and standard - deviation

The population mean $\mu = 1400$ and the standard deviation $\sigma=350$.
$\mu = 1400$, $\sigma = 350$

Step2: Calculate z - score

We want to find the z - score for $x = 1800$. The formula for the z - score is $z=\frac{x-\mu}{\sigma}$.
$z=\frac{1800 - 1400}{350}=\frac{400}{350}\approx1.14$

Step3: Find the percentage in the calculator

Using a standard normal distribution table or calculator, the area to the left of $z = 1.14$ is approximately $0.8729$ or $87.29\%$.

Step4: Graph and shade

Draw a normal - distribution curve. Mark the mean $\mu = 1400$ in the center. Mark the value $x = 1800$ to the right of the mean. Shade the area to the left of $x = 1800$.

Step1: Identify mean and standard - deviation

The population mean $\mu = 33$ and the standard deviation $\sigma = 4.5$.
$\mu = 33$, $\sigma = 4.5$

Step2: Calculate z - scores

For $x_1 = 30$:
$z_1=\frac{30 - 33}{4.5}=\frac{- 3}{4.5}\approx - 0.67$
For $x_2 = 40$:
$z_2=\frac{40 - 33}{4.5}=\frac{7}{4.5}\approx1.56$

Step3: Find the percentage in the calculator

The area to the left of $z_1=-0.67$ is approximately $0.2514$, and the area to the left of $z_2 = 1.56$ is approximately $0.9406$. The area between $z_1$ and $z_2$ is $0.9406-0.2514 = 0.6892$ or $68.92\%$.

Step4: Graph and shade

Draw a normal - distribution curve. Mark the mean $\mu = 33$ in the center. Mark $x_1 = 30$ to the left of the mean and $x_2 = 40$ to the right of the mean. Shade the area between $x_1 = 30$ and $x_2 = 40$.

Answer:

A. $\mu = 1400$, $\sigma = 350$
B. $z\approx1.14$
C. $87.29\%$
D. Graph with mean at 1400, mark 1800 to the right of the mean and shade area to the left of 1800

Question 18