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A small animal veterinarian reviews her records for the day and notes that she has seen eight dogs and eight cats with the following weights (in pounds).Dogs: 13, 25, 35, 49, 55, 65, 76, 105Cats: 3, 3, 6, 9, 11, 16, 18, 23a. Before analyzing these data sets, make a conjecture about which set has the larger mean, median, and standard deviation. Explain your reasoning.b. Compute the mean and standard deviation of each set.LLOROB. The mean, median, and standard deviation are all higher for cats because there is less variation in the weights, so the average value, middle value, andspread must be larger.c. The mean, median, and standard deviation are all higher for dogs because most of the weights are larger, so the average value, middle value, andspread must be larger.OD. The mean and median are higher for cats because there is less variation in the weights. The standard deviation is higher for dogs because there is morevariation in the weights.b. Compute the mean and standard deviation of each set.The mean for the dogs is(Type an integer or decimal rounded to one decimal place as needed.)27

User Xharze
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1 Answer

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10 votes

Given

eight dogs and eight cats with the following weights (in pounds).

Dogs: 13, 25, 35, 49, 55, 65, 76, 105

Cats: 3, 3, 6, 9, 11, 16, 18, 23

Find

a) make a conjecture about which set has the larger mean, median, and standard deviation. Explain your reasoning.

b) Compute the mean and standard deviation of each set.

Step-by-step explanation

a) The mean, median, and standard deviation are all higher for dogs because most of the weights are larger, so the average value, middle value, and spread must be larger.

b) mean of dogs =


\begin{gathered} (13+25+35+49+55+65+76+105)/(8) \\ \\ (423)/(8) \\ \\ 52.875\approx52.9 \end{gathered}

mean of cats =


\begin{gathered} (3+3+6+9+11+16+18+23)/(8) \\ \\ (89)/(8) \\ \\ 11.1 \end{gathered}

standard deviation of dogs =


\begin{gathered} \sigma=\sqrt{(1)/(N)\sum_{i\mathop{=}1}^N(x_i-\mu)^2} \\ \end{gathered}

on putting the values , we obtain


\sigma=29.5

standard deviation of cats =


\sigma=7.45\approx7.6

Final Answer

Therefore ,

mean

dogs : 52.9

Cats : 11.1

standard deviation

dogs: 29.5

cats: 7.6

User Bruria
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