The proportion of a normal distribution located between z = .50 and z = -.50 will be 38.2%.
We have,
A normal distribution located between z = 0.50 and z = -0.50,
So,
Now,
From the Z-score table,
We get,
The Probability corresponding to the Z score of -0.50,
i.e.
P(-0.50 < X < 0) = 0.191,
And,
The Probability corresponding to the Z score of -0.50,
i.e.
P(0 < X < 0.50) = 0.191,
Now,
The proportion of a normal distribution,
i.e.
P(Z₁ < X < Z₂) = P(Z₁ < X < 0) + P(0 < X < Z₂)
Now,
Putting values,
i.e.
P(-0.50 < X < 0.50) = P(-0.50 < X < 0) + P(0 < X < 0.50)
Now,
Again putting values,
We get,
P(-0.50 < X < 0.50) = 0.191 + 0.191
On solving we get,
P(-0.50 < X < 0.50) = 0.382
So,
We can write as,
P(-0.50 < X < 0.50) = 38.2%
So,
The proportion of a normal distribution is 38.2%.
Hence we can say that the proportion of a normal distribution located between z = .50 and z = -.50 will be 38.2%.
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