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:<br /><br />Mean (μ) = Σ(x * P(x))<br />Standard Deviation (σ) = √(Σ((x - μ)^2 * P(x)))<br /><br />Where x represents the temperature range, P(x) represents the probability of each temperature range, and Σ represents the sum of the products.<br /><br />Given data:<br />Temperature: $50-59$, $60-69$, $70-79$, $80-89$, $90-99$, $100-109$<br />Days: 4, 311, 1446, 1497, 516, 8<br /><br />Step 1: Calculate the mean (μ)<br />Mean (μ) = Σ(x * P(x))<br />Mean (μ) = ($50-59$ * 4/130) + ($60-69$ * 311/130) + ($70-79$ * 1446/130) + ($80-89$ * 1497/130) + ($90-99$ * 516/130) + ($100-109$ * 8/130)<br />Mean (μ) = 0.31 + 7.45 + 13.61 + 14.31 + 3.66 + 0.55<br />Mean (μ) = 39.9<br /><br />Step 2: Calculate the standard deviation (σ)<br />Standard Deviation (σ) = √(Σ((x - μ)^2 * P(x)))<br />Standard Deviation (σ) = √((($50-59$ - 39.9)^2 * 4/130) + (($60-69$ - 39.9)^2 * 311/130) + (($70-79$ - 39.9)^2 * 1446/130) + (($80-89$ - 39.9)^2 * 1497/130) + (($90-99$ - 39.9)^2 * 516/130) + (($100-109$ - 39.9)^2 * 8/130))<br />Standard Deviation (σ) = √(0.01 + 3.61 + 15.21 + 20.41 + 26.61 + 0.01)<br />Standard Deviation (σ) = 6.8<br /><br />Therefore, the approximate mean temperature is 39.9 and the standard deviation is 6.8.