1. **Problem Statement:** We have a frequency distribution of interview durations and need to calculate the mean and standard deviation, then interpret them. 2. **Given Data:** Class intervals and frequencies: - 10-20: 2 - 20-30: 4 - 30-40: 5 - 40-50: 8 - 50-60: 5 - 60-70: 1 Total frequency $n=25$ 3. **Calculate Midpoints ($x_i$):** - 10-20: $15$ - 20-30: $25$ - 30-40: $35$ - 40-50: $45$ - 50-60: $55$ - 60-70: $65$ 4. **Calculate Mean ($\bar{x}$):** Formula: $$\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$$ Calculate $\sum f_i x_i$: $$2\times15 + 4\times25 + 5\times35 + 8\times45 + 5\times55 + 1\times65 = 30 + 100 + 175 + 360 + 275 + 65 = 1005$$ Mean: $$\bar{x} = \frac{1005}{25} = 40.2$$ Interpretation: The average interview duration is approximately 40.2 minutes. 5. **Calculate Standard Deviation ($s$):** Formula: $$s = \sqrt{\frac{\sum f_i (x_i - \bar{x})^2}{n}}$$ Calculate each squared deviation times frequency: - $(15 - 40.2)^2 \times 2 = ( -25.2)^2 \times 2 = 635.04 \times 2 = 1270.08$ - $(25 - 40.2)^2 \times 4 = ( -15.2)^2 \times 4 = 231.04 \times 4 = 924.16$ - $(35 - 40.2)^2 \times 5 = ( -5.2)^2 \times 5 = 27.04 \times 5 = 135.2$ - $(45 - 40.2)^2 \times 8 = 4.8^2 \times 8 = 23.04 \times 8 = 184.32$ - $(55 - 40.2)^2 \times 5 = 14.8^2 \times 5 = 219.04 \times 5 = 1095.2$ - $(65 - 40.2)^2 \times 1 = 24.8^2 \times 1 = 615.04$ Sum: $$1270.08 + 924.16 + 135.2 + 184.32 + 1095.2 + 615.04 = 4224$$ Standard deviation: $$s = \sqrt{\frac{4224}{25}} = \sqrt{168.96} \approx 13.0$$ Interpretation: The interview durations vary on average by about 13 minutes from the mean. --- 6. **Section C: Milk Consumption Problem** Given: Mean $\mu=27.5$ kg, standard deviation $\sigma=6.5$ kg, normal distribution. 7. **(a) Probability consumed more than 15kg:** Calculate $Z$-score: $$Z = \frac{15 - 27.5}{6.5} = \frac{-12.5}{6.5} \approx -1.92$$ Probability: $$P(X > 15) = 1 - P(Z \leq -1.92) = 1 - 0.0274 = 0.9726$$ Interpretation: About 97.26% of people consume more than 15kg. 8. **(b) Probability consumed between 10kg and 20kg:** Calculate $Z$-scores: $$Z_{10} = \frac{10 - 27.5}{6.5} = -2.69$$ $$Z_{20} = \frac{20 - 27.5}{6.5} = -1.15$$ Probability: $$P(10 < X < 20) = P(-2.69 < Z < -1.15) = P(Z < -1.15) - P(Z < -2.69) = 0.1251 - 0.0036 = 0.1215$$ Interpretation: About 12.15% of people consume between 10kg and 20kg. 9. **(c) Probability consumed less than 10kg:** $$P(X < 10) = P(Z < -2.69) = 0.0036$$ Interpretation: About 0.36% of people consume less than 10kg. 10. **(d) Milk amount less than 99% of people consumed:** Find $Z$ for 99th percentile: $$Z = 2.33$$ Calculate milk amount: $$X = \mu + Z \sigma = 27.5 + 2.33 \times 6.5 = 27.5 + 15.145 = 42.645$$ Interpretation: 99% of people consume less than approximately 42.65 kg of milk.