Respuesta :
Answer:
(a) P (Z < 2.36) = 0.9909 (b) P (Z > 2.36) = 0.0091
(c) P (Z < -1.22) = 0.1112 (d) P (1.13 < Z > 3.35) = 0.1288
(e) P (-0.77< Z > -0.55) = 0.0705 (f) P (Z > 3) = 0.0014
(g) P (Z > -3.28) = 0.9995 (h) P (Z < 4.98) = 0.9999.
Step-by-step explanation:
Let us consider a random variable, [tex]X \sim N (\mu, \sigma^{2})[/tex], then [tex]Z=\frac{X-\mu}{\sigma}[/tex], is a standard normal variate with mean, E (Z) = 0 and Var (Z) = 1. That is, [tex]Z \sim N (0, 1)[/tex].
In statistics, a standardized score is the number of standard deviations an observation or data point is above the mean. The z-scores are standardized scores.
The distribution of these z-scores is known as the standard normal distribution.
(a)
Compute the value of P (Z < 2.36) as follows:
P (Z < 2.36) = 0.99086
≈ 0.9909
Thus, the value of P (Z < 2.36) is 0.9909.
(b)
Compute the value of P (Z > 2.36) as follows:
P (Z > 2.36) = 1 - P (Z < 2.36)
= 1 - 0.99086
= 0.00914
≈ 0.0091
Thus, the value of P (Z > 2.36) is 0.0091.
(c)
Compute the value of P (Z < -1.22) as follows:
P (Z < -1.22) = 0.11123
≈ 0.1112
Thus, the value of P (Z < -1.22) is 0.1112.
(d)
Compute the value of P (1.13 < Z > 3.35) as follows:
P (1.13 < Z > 3.35) = P (Z < 3.35) - P (Z < 1.13)
= 0.99960 - 0.87076
= 0.12884
≈ 0.1288
Thus, the value of P (1.13 < Z > 3.35) is 0.1288.
(e)
Compute the value of P (-0.77< Z > -0.55) as follows:
P (-0.77< Z > -0.55) = P (Z < -0.55) - P (Z < -0.77)
= 0.29116 - 0.22065
= 0.07051
≈ 0.0705
Thus, the value of P (-0.77< Z > -0.55) is 0.0705.
(f)
Compute the value of P (Z > 3) as follows:
P (Z > 3) = 1 - P (Z < 3)
= 1 - 0.99865
= 0.00135
≈ 0.0014
Thus, the value of P (Z > 3) is 0.0014.
(g)
Compute the value of P (Z > -3.28) as follows:
P (Z > -3.28) = P (Z < 3.28)
= 0.99948
≈ 0.9995
Thus, the value of P (Z > -3.28) is 0.9995.
(h)
Compute the value of P (Z < 4.98) as follows:
P (Z < 4.98) = 0.99999
≈ 0.9999
Thus, the value of P (Z < 4.98) is 0.9999.
**Use the z-table for the probabilities.
