Question

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

Answer 1

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.