1. **Problem Statement:** Calculate the mean lifetime and standard deviation of lifetimes for bulbs from manufacturers A and B based on the given grouped data. 2. **Formulas:** - Mean for grouped data: $$\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$$ where $f_i$ is frequency and $x_i$ is the midpoint of each class interval. - Standard deviation for grouped data: $$s = \sqrt{\frac{\sum f_i (x_i - \bar{x})^2}{\sum f_i}}$$ 3. **Calculate midpoints ($x_i$) for each class interval:** - 500–550: $\frac{500 + 550}{2} = 525$ - 550–600: $575$ - 600–650: $625$ - 650–700: $675$ - 700–750: $725$ - 750–800: $775$ - 800–850: $825$ - 850–900: $875$ 4. **Calculate mean for Manufacturer A:** - Frequencies $f_i$: 27, 50, 85, 99, 94, 74, 47, 24 - Calculate $\sum f_i x_i$: $$27\times525 + 50\times575 + 85\times625 + 99\times675 + 94\times725 + 74\times775 + 47\times825 + 24\times875$$ $$=14175 + 28750 + 53125 + 66825 + 68150 + 57350 + 38775 + 21000 = 347150$$ - Total bulbs $\sum f_i = 27 + 50 + 85 + 99 + 94 + 74 + 47 + 24 = 500$ - Mean $\bar{x}_A = \frac{347150}{500} = 694.3$ 5. **Calculate mean for Manufacturer B:** - Frequencies $f_i$: 22, 46, 78, 88, 92, 86, 52, 36 - Calculate $\sum f_i x_i$: $$22\times525 + 46\times575 + 78\times625 + 88\times675 + 92\times725 + 86\times775 + 52\times825 + 36\times875$$ $$=11550 + 26450 + 48750 + 59400 + 66700 + 66650 + 42900 + 31500 = 353900$$ - Total bulbs $\sum f_i = 22 + 46 + 78 + 88 + 92 + 86 + 52 + 36 = 500$ - Mean $\bar{x}_B = \frac{353900}{500} = 707.8$ 6. **Calculate standard deviation for Manufacturer A:** - Compute $\sum f_i (x_i - \bar{x}_A)^2$: \begin{align*} &(27)(525 - 694.3)^2 + (50)(575 - 694.3)^2 + (85)(625 - 694.3)^2 + (99)(675 - 694.3)^2 \\ &+ (94)(725 - 694.3)^2 + (74)(775 - 694.3)^2 + (47)(825 - 694.3)^2 + (24)(875 - 694.3)^2 \end{align*} - Calculate each squared difference: \begin{align*} &(27)(-169.3)^2 = 27 \times 28657.69 = 773760.63 \\ &(50)(-119.3)^2 = 50 \times 14232.49 = 711624.5 \\ &(85)(-69.3)^2 = 85 \times 4802.49 = 408211.65 \\ &(99)(-19.3)^2 = 99 \times 372.49 = 36876.51 \\ &(94)(30.7)^2 = 94 \times 942.49 = 88573.06 \\ &(74)(80.7)^2 = 74 \times 6512.49 = 481924.26 \\ &(47)(130.7)^2 = 47 \times 17082.49 = 802916.03 \\ &(24)(180.7)^2 = 24 \times 32652.49 = 783659.76 \end{align*} - Sum: $773760.63 + 711624.5 + 408211.65 + 36876.51 + 88573.06 + 481924.26 + 802916.03 + 783659.76 = 4003546.4$ - Standard deviation: $$s_A = \sqrt{\frac{4003546.4}{500}} = \sqrt{8007.09} = 89.49$$ 7. **Calculate standard deviation for Manufacturer B:** - Compute $\sum f_i (x_i - \bar{x}_B)^2$: \begin{align*} &(22)(525 - 707.8)^2 + (46)(575 - 707.8)^2 + (78)(625 - 707.8)^2 + (88)(675 - 707.8)^2 \\ &+ (92)(725 - 707.8)^2 + (86)(775 - 707.8)^2 + (52)(825 - 707.8)^2 + (36)(875 - 707.8)^2 \end{align*} - Calculate each squared difference: \begin{align*} &(22)(-182.8)^2 = 22 \times 33414.84 = 735126.48 \\ &(46)(-132.8)^2 = 46 \times 17644.84 = 811863.64 \\ &(78)(-82.8)^2 = 78 \times 6857.84 = 535410.72 \\ &(88)(-32.8)^2 = 88 \times 1075.84 = 94664.32 \\ &(92)(17.2)^2 = 92 \times 295.84 = 27192.32 \\ &(86)(67.2)^2 = 86 \times 4515.84 = 388362.24 \\ &(52)(117.2)^2 = 52 \times 13732.84 = 713109.68 \\ &(36)(167.2)^2 = 36 \times 27952.84 = 1006302.24 \end{align*} - Sum: $735126.48 + 811863.64 + 535410.72 + 94664.32 + 27192.32 + 388362.24 + 713109.68 + 1006302.24 = 4309931.64$ - Standard deviation: $$s_B = \sqrt{\frac{4309931.64}{500}} = \sqrt{8619.86} = 92.82$$ **Final answers:** - Mean lifetime Manufacturer A: $694.3$ days - Standard deviation Manufacturer A: $89.49$ days - Mean lifetime Manufacturer B: $707.8$ days - Standard deviation Manufacturer B: $92.82$ days