1. **State the problem:** We have a set of test scores out of 100, and we want to standardize these marks to better understand their distribution. 2. **List the scores:** 16, 0, 28, 12, 0, 78, 46, 0, 4, 12, 76, 8, 40, 10, 86, 76, 50, 32, 68, 40, 82, 16, 44, 34, 80, 44, 60, 70, 50, 0, 18 3. **Calculate the mean ($\mu$):** $$\mu = \frac{\sum x_i}{n} = \frac{16+0+28+12+0+78+46+0+4+12+76+8+40+10+86+76+50+32+68+40+82+16+44+34+80+44+60+70+50+0+18}{31}$$ Calculate the sum: $$16+0+28+12+0+78+46+0+4+12+76+8+40+10+86+76+50+32+68+40+82+16+44+34+80+44+60+70+50+0+18=1186$$ So, $$\mu = \frac{1186}{31} \approx 38.26$$ 4. **Calculate the standard deviation ($\sigma$):** $$\sigma = \sqrt{\frac{1}{n}\sum (x_i - \mu)^2}$$ Calculate each squared difference, sum them, then divide by 31 and take the square root. Sum of squared differences: $$\sum (x_i - 38.26)^2 = (16-38.26)^2 + (0-38.26)^2 + ... + (18-38.26)^2$$ Calculating each term and summing gives approximately 20488.52. So, $$\sigma = \sqrt{\frac{20488.52}{31}} = \sqrt{661.24} \approx 25.72$$ 5. **Standardize each score ($z$-score):** $$z = \frac{x - \mu}{\sigma}$$ For example, for 16: $$z = \frac{16 - 38.26}{25.72} \approx -0.86$$ 6. **Interpretation:** Each standardized score tells how many standard deviations the original score is from the mean. 7. **Final answer:** The mean is approximately $38.26$, the standard deviation is approximately $25.72$, and each score can be standardized using $$z = \frac{x - 38.26}{25.72}$$.