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
)
=
 .96

Draw picture and label it.

BlackTom AI · Accepted Answer

The random variable X is normally distributed with mean μ and standard deviation σ, so the standardized variable Z = (X − μ)/σ follows a standard normal distribution N(0,1).

We are asked to find z such that P(−z < (X − μ)/σ < z) = 0.96. Translating to Z, this is P(−z < Z < z) = 0.96.

Because the standard normal distribution is symmetric about zero, the probability in the interval (−z, z) is the difference between the CDF values at z and at −z: P(−z < Z < z) = Φ(z) − Φ(−z).

Using symmetry, Φ(−z) = 1 − Φ(z). Therefore, Φ(z) − Φ(−z) = Φ(z) − (1 − Φ(z)) = 2Φ(z) − 1.

Set 2Φ(z) − 1 = 0.96 and solve for Φ(z): 2Φ(z) = 1.96 ⇒ Φ(z) = 0.98.

Thus z must be the 0.98 quantile of the standard normal distribution. Consulting the standard normal table or using a calculator gives z ≈ 2.05 (more precisely about 2.0537, which rounds to 2.05 for two-decimal accuracy).

Intuitively, a 96% central probability around the mean leaves 2% in each tail, so z is the point where the cumulative probability reaches 0.98, captured by the standard normal 0.98 percentile.

BTRY3010/BTRY5010/STSCI2200/STSCI5200 Lecture #20 ( 92 - 93, 99 - 101 )

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 ) = .96 Draw picture and label it.

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