A normal distribution has mean μ and standard deviation σ. An x-value is randomly selected from the distribution. Find P (μ − 3σ ≤ x ≤ μ)

Respuesta :

Answer:

49.85%

Step-by-step explanation:

P(μ − 3σ ≤ x ≤ μ)

= ½ P(μ − 3σ ≤ x ≤ μ + 3σ)

= ½ (99.7%)

= 49.85%

Probability that a normal variate x lies in the range μ-3σ and μ is 0.4987.

How the area property of normal distribution works?

Area property of normal distribution as:

  • P(μ-σ<x<μ+σ) = 0.6826
  • P(μ-2σ<x<μ+2σ) = 0.9544
  • P(μ-3σ<x<μ+3σ) = 0.9973

To find P(μ-3σ<x<μ)

P(μ-3σ<x<μ) = 1/2 P(μ-3σ<x<μ+3σ)

=1/2 × 0.9973

=0.4987

Hence, the value of P(μ-3σ<x<μ) is 0.4987

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