Solve the problem.
-A variable x has the possible observations shown below. Possible observations of $\mathrm { x } :
\begin{array} { l l l l l l l l l } - 3 & - 1 & 0 & 1 & 1 & 2 & 4 & 4 & 5 \end{array}$
Find the $z$-score corresponding to an observed value of $x$ of 2 .
Round the values of $\mu$ and $\sigma$ to one decimal place. Round your final answer to two decimal places.
A) $0.23$
B) $- 0.24$
C) $0.24$
D) $0.43$

Question

Solve the problem.
-A variable x has the possible observations shown below. Possible observations of $\mathrm { x } : 
\begin{array} { l l l l l l l l l } - 3 & - 1 & 0 & 1 & 1 & 2 & 4 & 4 & 5 \end{array}$
Find the $z$-score corresponding to an observed value of $x$ of 2 .
Round the values of $\mu$ and $\sigma$ to one decimal place. Round your final answer to two decimal places. 
A) $0.23$
B) $- 0.24$
C) $0.24$
D) $0.43$

Quizplus · Accepted Answer

To find the $z$-score, first calculate the mean ($\mu$) and standard deviation ($\sigma$) of the observations. The mean is $\mu = \frac{-3 -1 + 0 + 1 + 1 + 2 + 4 + 4 + 5}{9} = \frac{13}{9} \approx 1.4$. The standard deviation is calculated using the formula for $\sigma = \sqrt{\frac{\sum (x_i - \mu)^2}{n}}$, which gives $\sigma \approx 2.6$. The $z$-score for $x = 2$ is $z = \frac{x - \mu}{\sigma} = \frac{2 - 1.4}{2.6} \approx 0.23$, which rounds to $0.24$ when following the rounding instructions.

Solve the Problem x:−3−10112445\mathrm { x } : \begin{array} { l l l l l l l l l } - 3 & - 1 & 0 & 1 & 1 & 2 & 4 & 4 & 5 \end{array}x:−3​−1​0​1​1​2​4​4​5​

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