Question

If a normal distribution has a mean of 62 and a standard deviation of 12 what is the z-score for a value of 80

Answer 1

Answer with explanation:

$\text{Mean}=\mu = 62\\\\ \text{standard deviation}=\sigma=12\\\\ \bar X=80\\\\Z_{score}=\frac{\bar X -\mu}{\sigma}\\\\Z_{score}=\frac{80-62}{12}\\\\Z_{score}=\frac{18}{12}\\\\Z_{score}=1.5$

Option B:→ 1.5