The Practice of Statistics for AP - 4th Edition
The Practice of Statistics for AP - 4th Edition
4th Edition
ISBN: 9781429245593
Author: Starnes, Daren S., Yates, Daniel S., Moore, David S.
Publisher: Macmillan Higher Education
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Chapter 2.2, Problem 55E

(a)

To determine

To find: the proportion of days where the given condition falls

(a)

Expert Solution
Check Mark

Answer to Problem 55E

Therefore, we would expect trains to arrive on time about 96% of the time.

Explanation of Solution

Given:

Mean is μ=0.37

Standard deviation is σ=0.04 .

Calculation:

It is given that for a certain small train's daily route, the locomotive needs to have anadhesion of at least 0.30 for the train at its destination on time. We want to find the area under the N(0.37, 0.04) distribution to the right of 0.3.

We first calculate the z-value corresponding to 0.3.

  z=0.30.370.04=1.75

The area above -1.75 is equal to the area to the right of 0.3. The following graph shows the area above 0.3.

  The Practice of Statistics for AP - 4th Edition, Chapter 2.2, Problem 55E , additional homework tip  1

From standard normal table, the value below -1.75 is 0.0401. Then, the area above -1.75 is

1- 0.0401 = 0.9599.

Therefore, it is expectedtrains to arrive on time about 96% of the time.

Conclusion:

Therefore, it is expected trains to arrive on time about 96% of the time.

(b)

To determine

To find: the proportion of days where the given condition falls

(b)

Expert Solution
Check Mark

Answer to Problem 55E

It is expectedtrains to arrive early about 0.06% of the time.

Explanation of Solution

Calculation:

We first find the z -value corresponding to 0.5 as follows:

  z=0.50.370.04=3.25

The area above 0.5 is equal to the area to the right of 3.25. The following graph shows the area above 0.5.

  The Practice of Statistics for AP - 4th Edition, Chapter 2.2, Problem 55E , additional homework tip  2

Using standard normal table, the proportion of adhesions higher than 0.50 is1-0.9994 = 0.0006.

Hence, it is expectedtrains to arrive early about 0.06% of the time.

Conclusion:

Hence, it is expectedtrains to arrive early about 0.06% of the time.

(c)

To determine

To compare: the answers to (a) and (b).

(c)

Expert Solution
Check Mark

Answer to Problem 55E

It makes sense to try to have the value found in part (a) larger.

Explanation of Solution

Calculation:

It makes sense to try to have the value found in part (a) larger. It is expected to have the train which is to arrive atits destination on time, but not to arrive at the switch point early.

Conclusion:

It makes sense to try to have the value found in part (a) larger.

Chapter 2 Solutions

The Practice of Statistics for AP - 4th Edition

Ch. 2.1 - Prob. 4.1CYUCh. 2.1 - Prob. 4.2CYUCh. 2.1 - Prob. 4.3CYUCh. 2.1 - Prob. 1ECh. 2.1 - Prob. 2ECh. 2.1 - Prob. 3ECh. 2.1 - Prob. 4ECh. 2.1 - Prob. 5ECh. 2.1 - Prob. 6ECh. 2.1 - Prob. 7ECh. 2.1 - Prob. 8ECh. 2.1 - Prob. 9ECh. 2.1 - Prob. 10ECh. 2.1 - Prob. 11ECh. 2.1 - Prob. 12ECh. 2.1 - Prob. 13ECh. 2.1 - Prob. 14ECh. 2.1 - Prob. 15ECh. 2.1 - Prob. 16ECh. 2.1 - Prob. 17ECh. 2.1 - Prob. 18ECh. 2.1 - Prob. 19ECh. 2.1 - Prob. 20ECh. 2.1 - Prob. 21ECh. 2.1 - Prob. 22ECh. 2.1 - Prob. 23ECh. 2.1 - Prob. 24ECh. 2.1 - Prob. 25ECh. 2.1 - Prob. 26ECh. 2.1 - Prob. 27ECh. 2.1 - Prob. 28ECh. 2.1 - Prob. 29ECh. 2.1 - Prob. 30ECh. 2.1 - Prob. 31ECh. 2.1 - Prob. 32ECh. 2.1 - Prob. 33ECh. 2.1 - Prob. 34ECh. 2.1 - Prob. 35ECh. 2.1 - Prob. 36ECh. 2.1 - Prob. 37ECh. 2.1 - Prob. 38ECh. 2.1 - Prob. 39ECh. 2.1 - Prob. 40ECh. 2.2 - Prob. 1.1CYUCh. 2.2 - Prob. 1.2CYUCh. 2.2 - Prob. 1.3CYUCh. 2.2 - Prob. 2.1CYUCh. 2.2 - Prob. 2.2CYUCh. 2.2 - Prob. 2.3CYUCh. 2.2 - Prob. 2.4CYUCh. 2.2 - Prob. 2.5CYUCh. 2.2 - Prob. 3.1CYUCh. 2.2 - Prob. 3.2CYUCh. 2.2 - Prob. 3.3CYUCh. 2.2 - Prob. 41ECh. 2.2 - Prob. 42ECh. 2.2 - Prob. 43ECh. 2.2 - Prob. 44ECh. 2.2 - Prob. 45ECh. 2.2 - Prob. 46ECh. 2.2 - Prob. 47ECh. 2.2 - Prob. 48ECh. 2.2 - Prob. 49ECh. 2.2 - Prob. 50ECh. 2.2 - Prob. 51ECh. 2.2 - Prob. 52ECh. 2.2 - Prob. 53ECh. 2.2 - Prob. 54ECh. 2.2 - Prob. 55ECh. 2.2 - Prob. 56ECh. 2.2 - Prob. 57ECh. 2.2 - Prob. 58ECh. 2.2 - Prob. 59ECh. 2.2 - Prob. 60ECh. 2.2 - Prob. 61ECh. 2.2 - Prob. 62ECh. 2.2 - Prob. 63ECh. 2.2 - Prob. 64ECh. 2.2 - Prob. 65ECh. 2.2 - Prob. 66ECh. 2.2 - Prob. 67ECh. 2.2 - Prob. 68ECh. 2.2 - Prob. 69ECh. 2.2 - Prob. 70ECh. 2.2 - Prob. 71ECh. 2.2 - Prob. 72ECh. 2.2 - Prob. 73ECh. 2.2 - Prob. 74ECh. 2.2 - Prob. 75ECh. 2.2 - Prob. 76ECh. 2 - Prob. 1CRECh. 2 - Prob. 2CRECh. 2 - Prob. 3CRECh. 2 - Prob. 4CRECh. 2 - Prob. 5CRECh. 2 - Prob. 6CRECh. 2 - Prob. 7CRECh. 2 - Prob. 8CRECh. 2 - Prob. 9CRECh. 2 - Prob. 10CRECh. 2 - Prob. 11CRECh. 2 - Prob. 12CRECh. 2 - Prob. 1PTCh. 2 - Prob. 2PTCh. 2 - Prob. 3PTCh. 2 - Prob. 4PTCh. 2 - Prob. 5PTCh. 2 - Prob. 6PTCh. 2 - Prob. 7PTCh. 2 - Prob. 8PTCh. 2 - Prob. 9PTCh. 2 - Prob. 10PTCh. 2 - Prob. 11PTCh. 2 - Prob. 12PTCh. 2 - Prob. 13PT
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