Suppose X is a random variable with a normal distribution with mean mu and standard deviation sigma. What is the value of z that makes this a true statement? ( Find Z that satisfies this equation )

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Question

Suppose X is a random variable with a normal distribution with mean mu and standard deviation sigma. What is the value of  z that makes this a true statement? ( Find Z that satisfies this equation )

𝑃
(
−
𝑧
<
𝑋
−
𝜇
𝜎
<
𝑧
)
=
 .90

Draw picture and label it.

BlackTom AI · Accepted Answer

The problem describes a standard-normal transformation to compare a probability interval to a z-value. The random variable X ~ N(μ, σ^2) can be standardized by Z = (X − μ)/σ, which follows N(0, 1).

We are asked to find z such that P(−z < (X − μ)/σ < z) = 0.90. In terms of Z, this is P(−z < Z < z) = 0.90, which is the probability that Z lies within ±z around 0.

Because the standard normal distribution is symmetric about 0, the central interval (−z, z) captures 0.90 of the probability, leaving 0.10 in the two tails combined. By symmetry, each tail has 0.05.

Thus, the interval must extend to the point where the cumulative distribution function satisfies P(Z ≤ z) = 0.95, i.e., z is the 0.95-quantile (the 95th percentile) of the standard normal distribution. In other words, z = z_{0.95}.

From standard normal tables or a calculator, the value of z_{0.95} is approximately 1.645. This means the central region between −1.645 and 1.645 contains about 90% of the probability.

To visualize, imagine the normal curve with the central band from −z to z shaded; the shaded region should cover 90% of the area under the curve, with 5% in each tail beyond ±z.

Common pitfalls include misinterpreting the two-tailed requirement or mixing up the standardization step. Remember, you’re looking for the z such that the area between −z and z equals 0.90, which directly ties to the 95th percentile of the standard normal when using symmetry.

BTRY3010/BTRY5010/STSCI2200/STSCI5200 Lecture#21 (Pages 102-103)

Suppose X is a random variable with a normal distribution with mean mu and standard deviation sigma. What is the value of z that makes this a true statement? ( Find Z that satisfies this equation ) 𝑃 ( − 𝑧 < 𝑋 − 𝜇 𝜎 < 𝑧 ) = .90 Draw picture and label it.

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