# Demonstrate probability determined

1) Tire Life - The life of a tire is normally distributed with a mean of 76,000 miles and a standard deviation of 10,000 miles.

a) Determine the probability that a tire will last for less than 67,000 miles.

b) Determine the probability that a tire will last for more than 67,000 miles.

c) Determine the probability that a tire will last for less than 80,000 miles.

d) Determine the probability that a tire will last between 67000 and 80000 miles?

e) The company offers a warranty. The warranty costs $8 per tire returned. If the company sets the warranty miles at 67000, how much on average per tire will the warranty cost?

2) Church Fund Raiser - The amount of money earned at the church charity auction for the last 20 years is in the following table. Assume a normal distribution. (Use Excel)

What is the mean?

Year Amount

1 349

2 280

3 390

4 470

5 380

6 430

7 290

8 477

9 298

10 449

11 370

12 375

13 405

14 490

15 402

16 410

17 330

18 510

19 325

20 370

What is the standard deviation?

What is the probability that the church will earn less than $425 next year?

What is the probability that the church will earn more than $425 next year?

What is the probability that the amount will be between $400 and $475?

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#### Solution Preview

I will transfer the entire work sheet to excel (Q1 + Q2). All explanations will be provided here.

1. There is a function in excel called NORM.DIST(x, mean, standard_dev ,cumulative) gives the probability that a number falls at or below a given value ...

#### Solution Summary

The expert demonstrates probability determined. The probability that a tire will last 67,000 miles is determined.