Answered

Suppose $X \sim N(16.5,0.5)$, and $x=16$. Find and interpret the $z$-score of the standardized normal random variable.



The $z$-score when $x=16$ is [___]. The mean is [___].



This $z$-score tells you that $x=16$ is [___] standard deviation(s) to the left of the mean.

Verified Answer :

  • Identify the parameters: \(\mu = 16.5\), \(\sigma = 0.5\), and \(x = 16\).
  • Apply the formula: \(z = \frac{x - \mu}{\sigma} = \frac{16 - 16.5}{0.5} = -1\).
  • Interpret the result: The value is 1 standard deviation to the left of the mean.
  • The \(z\)-score when \(x = 16\) is \(\boxed{-1}\). The mean is \(\boxed{16.5}\). This \(z\)-score tells you that \(x = 16\) is \(\boxed{1}\) standard deviation(s) to the left of the mean.

Explanation

  1. Understand the components of the z-score formula

The problem involves a normally distributed random variable \(X \sim N(\mu, \sigma)\) with a mean \(\mu = 16.5\) and a standard deviation \(\sigma = 0.5\). We are asked to find the \(z\)-score for a specific observation \(x = 16\) and interpret this value relative to the mean. The \(z\)-score measures how many standard deviations a data point is from the mean of a population. It is calculated using the formula:
\[z = \frac{x - \mu}{\sigma}\]

  1. Calculate the z-score

Substituting the given values \(x = 16\), \(\mu = 16.5\), and \(\sigma = 0.5\) into the \(z\)-score formula:
\[z = \frac{16 - 16.5}{0.5} = \frac{-0.5}{0.5} = -1.0\]
The \(z\)-score is \(-1.0\).

  1. Interpret the result

The \(z\)-score of \(-1.0\) indicates that the value \(x = 16\) is exactly \(1\) standard deviation below (to the left of) the mean of \(16.5\). The negative sign specifically indicates the position to the left of the mean, and the magnitude of \(1\) indicates the distance in standard deviations.

The \(z\)-score when \(x = 16\) is \(-1\). The mean is \(16.5\).

This \(z\)-score tells you that \(x = 16\) is \(1\) standard deviation(s) to the left of the mean.

Examples

In quality control, this method is used to determine how far a specific product measurement deviates from the process mean. For example, if a machine is set to produce parts with a length of 16.5 cm with a standard deviation of 0.5 cm, finding a part with a length of 16 cm (a z-score of -1) helps engineers understand if the process is within expected performance limits or if it requires adjustment.

Related Questions