Find the following probabilities for the standard normal random variable z: a. P(−1.36 ≤ z ≤ 1.38)= ________ b. P(−2.27 ≤ z ≤ 1.64)= ________ c. P(z ≤ 1.33)=________ d. P(z > −0.68)=_____

Respuesta :

Answer:

Step-by-step explanation:

a) P(−1.36 ≤ z ≤ 1.38)

For z = - 1.36,

The corresponding probability from the normal distribution table is 0.08691

For z = 1.38

The corresponding probability from the normal distribution table is 0.9162

Therefore,

P(−1.36 ≤ z ≤ 1.38)

= 0.9162 - 0.08691 = 0.82929

b) P(−2.27 ≤ z ≤ 1.64)

For z = - 2.27

The corresponding probability from the normal distribution table is

0.0116

For z = 1.64,

The corresponding probability from the normal distribution table is

0.9495

Therefore,

P(−2.27 ≤ z ≤ 1.64)

= 0.9495 - 0.0116 = 0.9379

c) P(z ≤ 1.33)

For z = 1.33

The corresponding probability from the normal distribution is 0.9082

Therefore,

P(z ≤ 1.33) = 0.9082

d) P(z > −0.68) = 1 - P(z ≤ −0.68)

For z = - 0.68

The corresponding probability from the normal distribution is 0.2482

Therefore,

P(z > −0.68) = 1 - 0.2482 = 0.7518

Answer:

a) 0.8293

b) 0.9379

c) 0.9082

d) 0.7517

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

a. P(−1.36 ≤ z ≤ 1.38)=

This is the pvalue of Z = 1.38 subtracted by the pvalue of Z = -1.36

Z = 1.38 has a pvalue of 0.9162

Z = -1.36 has a pvalue of 0.0869

0.9162 - 0.0869 = 0.8293

b. P(−2.27 ≤ z ≤ 1.64)

Same logic as a.

Z = 1.64 has a pvalue of 0.9495

Z = -2.27 has a pvalue of 0.0116

0.9495 - 0.0116 = 0.9379

c. P(z ≤ 1.33)=

This is the pvalue of Z = 1.33.

Z = 1.33 has a pvalue of 0.9082

d. P(z > −0.68)=

This is 1 subtracted by the pvalue of Z = -0.68. So

Z = -0.68 has a pvalue of 0.2483

1 - 0.2483 = 0.7517

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