Warm-Up Honors Algebra 2 3/27/19

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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?

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 .4357.

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

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

Find the area to the right of z = 1.11 Answer: whole is 1. Look up z = 1.11 to get an area of 0.8665. Final Answer: 1 – 0.8665 = 0.1335 0.8665 1.0

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.

Find the area between z = 2.00 and z = 2.47. Look up z = 2.47 to get 0.9932. Look up z = 2 to get 0.9772. Subtract the two areas: 0.9932 – 0.9772 = 0.0160 How can we do this? What should I do first? (Draw & Shade)

You try…… Find the area between z = -2.48 and z = -0.83. Answer: 0.2033 – 0.0068 = 0.1365

Find the area between z = -1.37 and z = 1.68. The area for z = 1.68 is 0.9535. The area for z = =1.37 is 0.0853. Add the two areas together for the final answer: 0.9535 + 0.0853 = 0.8682

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

Find the area to the right of z = -1.16. The area for z = -1.16 is 0.1230. The area for the entire distribution is 1.00. Subtract the areas : 1 - 0.1230 = 0.8770

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 – 0.9925 = 0.0075. The area for z = -3.01 is 0.0013. Add the two areas together: 0.0075 + 0.0013 = 0.0088