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Assume that z-scores are normally distributed with a mean of 0 and a standard deviation of If P(−d

User Bircan
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In a standard normal distribution with mean 0 and standard deviation 1, the value of
\(a\) for
\(P(-a < z < a) = 0.4314\) is approximately
\(a = 0.2157\), encompassing 43.14% of the distribution's area between
\(-a\) and
\(a\).

Given that
\( z \) scores are normally distributed with a mean of 0 and a standard deviation of 1, and the probability
\( P(-a < z < a) = 0.4314 , we'll use the properties of the standard normal distribution to find
\( a \).

For a standard normal distribution, the area between
\( -a \) and
\( a \) under the curve represents the probability
\( P(-a < z < a) \).

Given
\( P(-a < z < a) = 0.4314 \), we want to find the value of
\( a \).

From the properties of symmetry in the standard normal distribution, the total area between
\( -a \) and
\( a \) is
\( 2a \) , which corresponds to the probability
\( P(-a < z < a) \).

Given:


\[ P(-a < z < a) = 0.4314 \]

Since the total area is
\( 2a \), and
\( P(-a < z < a) = 0.4314 \):


\[ 2a = 0.4314 \]


\[ a = (0.4314)/(2) \]


\[ a = 0.2157 \]

Therefore, the value of
\( a \) such that
\( P(-a < z < a) = 0.4314 \) in a standard normal distribution is approximately
\( a = 0.2157 \).

complete the question

Assume that z scores are normally distributed with a mean of 0 and a standard deviation of 1. If P (-a < z < a) = 0.4314, find a.

User Gone
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