The following data represent the high temperature distribution for a summer month in a city for some of the last 130 years. Treat the data as a population. Complete parts (a) through (c) Temperature 50-59 60-69 70-79 80-89 90-99 100-109 Davs 4 311 1446 1497 516 8 (a) Approximate the mean and standard deviation for temperature. mu =square (Round to one decimal place as needed.)

To approximate the mean and standard deviation for temperature, we can use the following formulas:Mean (μ) = Σ(x * P(x))Standard Deviation (σ) = √(Σ((x - μ)^2 * P(x)))Where x represents the temperature range, P(x) represents the probability of each temperature range, and Σ represents the sum of the products.Given data:Temperature: , , , , , Days: 4, 311, 1446, 1497, 516, 8Step 1: Calculate the mean (μ)Mean (μ) = Σ(x * P(x))Mean (μ) = ( * 4/130) + ( * 311/130) + ( * 1446/130) + ( * 1497/130) + ( * 516/130) + ( * 8/130)Mean (μ) = 0.31 + 7.45 + 13.61 + 14.31 + 3.66 + 0.55Mean (μ) = 39.9Step 2: Calculate the standard deviation (σ)Standard Deviation (σ) = √(Σ((x - μ)^2 * P(x)))Standard Deviation (σ) = √((( - 39.9)^2 * 4/130) + (( - 39.9)^2 * 311/130) + (( - 39.9)^2 * 1446/130) + (( - 39.9)^2 * 1497/130) + (( - 39.9)^2 * 516/130) + (( - 39.9)^2 * 8/130))Standard Deviation (σ) = √(0.01 + 3.61 + 15.21 + 20.41 + 26.61 + 0.01)Standard Deviation (σ) = 6.8Therefore, the approximate mean temperature is 39.9 and the standard deviation is 6.8.