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The area beneath the normal density curve is separated into three regions by the values
and
. Set the mean of the normal density curve to 6.4, the standard deviation to 0.7, and the -value
of
to 6.7. First, determine the value of the area to the right of .

Place
so that it equally divides the area to the left of
. What is the -value
of ?


Sagot :

fichoh

The value of the area to the right of X > 6.7 calculated from the normal distribution principle is 0.334

Given the Parameters :

  • Mean, μ = 6.4
  • Standard deviation, σ = 0.7
  • X = 6.7

Using the Zscore formula for a normal distribution :

  • Zscore = (X - mean) / standard deviation

The Zscore, when X = 6.7

  • Zscore = (6.7 - 6.4) / 0.7 = 0.4286

The area to the right of the distribution :

  • P(Z > 0.4286) = 1 - P(Z < 0.4286)

Using a normal distribution table or calculator :

  • P(Z > 0.4286) = 1 - 0.66589 = 0.334

Therefore, the probability value of P(X < 6.7) is

  • P(Z > 0.4286) = 0.334

Learn more :https://brainly.com/question/8165716

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