1. Warm-Up Honors Algebra 2 3/27/19
The length of time it takes a teenager to respond to a Facebook update is (surprisingly) normally distributed with a mean of 4 hours and standard deviation of 23 minutes. What is the probability that a recorded time will not be in between 217 and 286 minutes?

2. Example……Find the area under the curve between z = 0 and z = 1.52.
Look up z = 1.52
The area between z = 0 and z = 1.52 is

3. You try…Find the area between z = 0 and z = 2.34.
Answer:
.4904

4. Find the area between z = 0 and z = -1.75.
Answer:
.4599

5. Find the area to the right of z = 1.11
Answer:
whole is 1.
Look up z = 1.11 to get an area of
Final Answer:
1 – =
0.8665
1.0

6. Now you try…… Find the area to the left of z = -1.93. Answer: 0.0268
Compare results with a neighbor, I walk around to keep on task.

7. Find the area between z = 2.00 and z = 2.47.
Look up z = 2.47 to get
Look up z = 2 to get
Subtract the two areas:
– =
How can we do this? What should I do first? (Draw & Shade)

8. You try…… Find the area between z = -2.48 and z = -0.83.
Answer: – =

9. Find the area between z = -1.37 and z = 1.68.
The area for z = 1.68 is
The area for z = =1.37 is
Add the two areas together for the final answer: =

10. Find the area to the left of z = 1.99.
The area for z = 1.99 is
0.9767

11. Find the area to the right of z = -1.16.
The area for z = is
The area for the entire distribution is 1.00.
Subtract the areas : =

12. Find the area to the right of z = 2.43 and to the left of z = -3.01.
The area for z = 2.43 is 1 – =
The area for z = is
Add the two areas together: =