If degree of freedom(df) = 20, find the t value(s) for α = 1.0% in two tails(lower and upper tail) ✡✎ +/-2.845+/-2.145+2.861-2.861If degree of freedom(df) = 2, find the t value for α = 5.0% in one tail( lower tail) ✡✎ -2.920+2.920-1.7532.977none of the aboveIf degree of freedom(df) = 20, find the t value for α = 1.0% in one tail( lower tail)– 2.528 ✢✎ ☞ ✒✎✕✒✘-2.539+ 2.539 If n = 20, find the t value for α = 0.5% in one tail (upper tail)2.8612.602 ✣✎ 2.9772.947none of the above Use the standard normal probability distribution table (Z) to find the following probabilities:P( Z < 1.2)5871547865438849P( Z > 1.72)0427957396560344P( Z < -0.42)0778922233726638P( Z > - 0.42)0778922233226628P( 2.28 < Z <3.28)0778999501089887 P( -1.20 < Z < 1 .04)8508735774579876 P( -1.52 < Z < -0.09)098706430,39984641

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Calculus For The Life Sciences

2nd Edition

ISBN:9780321964038

Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.

Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.

Chapter2: Exponential, Logarithmic, And Trigonometric Functions

Section2.CR: Chapter 2 Review

Problem 111CR: Respiratory Rate Researchers have found that the 95 th percentile the value at which 95% of the data...

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Question

If degree of freedom(df) = 20, find the t value(s) for α = 1.0% in two tails(lower and upper tail)

✡✎ +/-2.845

+/-2.145
+2.861
-2.861
If degree of freedom(df) = 2, find the t value for α = 5.0% in one tail( lower tail)

✡✎ -2.920

+2.920
-1.753
2.977
none of the above
If degree of freedom(df) = 20, find the t value for α = 1.0% in one tail( lower tail)
– 2.528

✢✎ ☞ ✒✎✕✒✘

-2.539
+ 2.539

If n = 20, find the t value for α = 0.5% in one tail (upper tail)
2.861
2.602

✣✎ 2.977

2.947
none of the above

Use the standard normal probability distribution table (Z) to find the following probabilities:

P( Z < 1.2)
1. 5871
2. 5478
3. 6543
4. 8849
P( Z > 1.72)
1. 0427
2. 9573
3. 9656
4. 0344
P( Z < -0.42)
1. 0778
2. 9222
3. 3372
4. 6638
P( Z > - 0.42)
1. 0778
2. 9222
3. 3322
4. 6628
P( 2.28 < Z <3.28)
1. 0778
2. 9995
3. 0108
4. 9887

P( -1.20 < Z < 1 .04)
1. 8508
2. 7357
3. 7457
4. 9876

P( -1.52 < Z < -0.09)
1. 0987
2. 0643
3. 0,3998
4. 4641

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

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