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For a standard normal distribution, which of the following expressions must always be equal to 1?
A) P(z≤-a)-P(-a≤z≤a)+P(z≥a)
B) P(z≤-a)-P(-a≤z≤a)+P(z≥a)
C) P(z≤-a)+P(-a≤z≤a)-P(z≥a)
D) P(z≤-a)+P(-a≤z≤a)+P(Z≥a)

Respuesta :

nobillionaireNobley
P(z ≤ -a) + P(-a ≤ z ≤ a) + P(z ≥ a) = 1 - P(z ≤ a) + [P(z ≤ a) - P(z ≤ -a)] + 1 - P(z ≤ a) = 2 - 2P(z ≤ a) + P(z ≤ a) - [1 - P(z ≤ a)] = 2 - P(z ≤ a) - 1 + P(z ≤ a) = 1

Therefore, option D is the correct answer.
InesWalston

Answer:

D. [tex]P(z\le -a)+P(-a\le z\le a)+P(z\ge a)[/tex]

Step-by-step explanation:

Properties of normal distribution-

  1. The normal curve is symmetrical about the mean (μ).
  2. The mean is at the middle of the graph and it divides the area into two equal halves.
  3. The total area under the curve is equal to 1.

The total area under the curve can be divided into parts like,

  1. area below [tex]-a[/tex], i.e [tex]z\le -a[/tex],
  2. area between [tex]-a[/tex] to [tex]a[/tex], i.e [tex]-a\le z\le a[/tex]
  3. area above [tex]a[/tex], i.e [tex]z\ge a[/tex]

Therefore, [tex]P(z\le -a)+P(-a\le z\le a)+P(z\ge a)=1[/tex]

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