Question

question 24 (bonus) (1 point)
use the standard normal distribution table to help answer the question below.
scores are normally distributed with a mean of 31 and standard deviation of 4.
what is the approximate percentage of times we observe a score above 39?
______%
round the z - score to two decimal places if needed. give answer to two decimal
places after converting to a percent. do not include the \%\ sign in your answer.
your answer:
answer
question 25 (bonus) (1 point)
a frozen pizza manufacturer needs to make 9 inch pizzas. their machine that makes
the dough is set to make a pizza with a diameter of 9.1 inches, but it is known that
the distribution of pizza diameters has roughly a normal distribution with a mean of
the setting of 9.1 inches and a standard deviation of 0.06 inches. if people measure
their pizza and get less than 9 inches they may complain. find the percentage of
pizzas that are less than 9 inches based on the given information.
_____ percent of pizzas that are made below 9 inches.
round the z - score to two decimal places if needed. round your answer to two
decimal places after converting to a percent. do not include the % sign in your
answer.

Explanation:

Question 24

Step1: Calculate the z - score

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(x = 39\), \(\mu=31\), and \(\sigma = 4\).

$$z=\frac{39 - 31}{4}=\frac{8}{4}=2.00$$

Step2: Find the probability

We want \(P(X>39)\), which is equivalent to \(P(Z > 2.00)\) (by the z - score transformation). Since the total area under the standard normal curve is \(1\), and \(P(Z\leq z)+P(Z > z)=1\). From the standard normal distribution table, \(P(Z\leq2.00)=0.9772\).

$$P(Z > 2.00)=1 - P(Z\leq2.00)=1-0.9772 = 0.0228$$

Step1: Calculate the z - score

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(x = 9\), \(\mu = 9.1\), and \(\sigma=0.06\)

$$z=\frac{9 - 9.1}{0.06}=\frac{- 0.1}{0.06}\approx - 1.67$$

Step2: Find the probability

We want \(P(X < 9)\), which is equivalent to \(P(Z<-1.67)\) (by the z - score transformation). From the standard normal distribution table, \(P(Z < - 1. 67)=0.047 5\)

Answer:

\(2.28\)

Question 25