State whether the equation represents exponential growth, exponential decay, or neither. y=2*0.68^x
2 answers:
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Answer:
(a) P (<em>Z</em> < 2.36) = 0.9909 (b) P (<em>Z</em> > 2.36) = 0.0091
(c) P (<em>Z</em> < -1.22) = 0.1112 (d) P (1.13 < <em>Z</em> > 3.35) = 0.1288
(e) P (-0.77< <em>Z</em> > -0.55) = 0.0705 (f) P (<em>Z</em> > 3) = 0.0014
(g) P (<em>Z</em> > -3.28) = 0.9995 (h) P (<em>Z</em> < 4.98) = 0.9999.
Step-by-step explanation:
Let us consider a random variable, , then
, is a standard normal variate with mean, E (<em>Z</em>) = 0 and Var (<em>Z</em>) = 1. That is,
.
In statistics, a standardized score is the number of standard deviations an observation or data point is above the mean. The <em>z</em>-scores are standardized scores.
The distribution of these <em>z</em>-scores is known as the standard normal distribution.
(a)
Compute the value of P (<em>Z</em> < 2.36) as follows:
P (<em>Z</em> < 2.36) = 0.99086
≈ 0.9909
Thus, the value of P (<em>Z</em> < 2.36) is 0.9909.
(b)
Compute the value of P (<em>Z</em> > 2.36) as follows:
P (<em>Z</em> > 2.36) = 1 - P (<em>Z</em> < 2.36)
= 1 - 0.99086
= 0.00914
≈ 0.0091
Thus, the value of P (<em>Z</em> > 2.36) is 0.0091.
(c)
Compute the value of P (<em>Z</em> < -1.22) as follows:
P (<em>Z</em> < -1.22) = 0.11123
≈ 0.1112
Thus, the value of P (<em>Z</em> < -1.22) is 0.1112.
(d)
Compute the value of P (1.13 < <em>Z</em> > 3.35) as follows:
P (1.13 < <em>Z</em> > 3.35) = P (<em>Z</em> < 3.35) - P (<em>Z</em> < 1.13)
= 0.99960 - 0.87076
= 0.12884
≈ 0.1288
Thus, the value of P (1.13 < <em>Z</em> > 3.35) is 0.1288.
(e)
Compute the value of P (-0.77< <em>Z</em> > -0.55) as follows:
P (-0.77< <em>Z</em> > -0.55) = P (<em>Z</em> < -0.55) - P (<em>Z</em> < -0.77)
= 0.29116 - 0.22065
= 0.07051
≈ 0.0705
Thus, the value of P (-0.77< <em>Z</em> > -0.55) is 0.0705.
(f)
Compute the value of P (<em>Z</em> > 3) as follows:
P (<em>Z</em> > 3) = 1 - P (<em>Z</em> < 3)
= 1 - 0.99865
= 0.00135
≈ 0.0014
Thus, the value of P (<em>Z</em> > 3) is 0.0014.
(g)
Compute the value of P (<em>Z</em> > -3.28) as follows:
P (<em>Z</em> > -3.28) = P (<em>Z</em> < 3.28)
= 0.99948
≈ 0.9995
Thus, the value of P (<em>Z</em> > -3.28) is 0.9995.
(h)
Compute the value of P (<em>Z</em> < 4.98) as follows:
P (<em>Z</em> < 4.98) = 0.99999
≈ 0.9999
Thus, the value of P (<em>Z</em> < 4.98) is 0.9999.
**Use the <em>z</em>-table for the probabilities.