In December 2024
Answer to Question #152049 in Statistics and Probability for JG
a) If we assume the data follows the normal distribution, for a randomly received emergency call, what is the probability that the response will be above 13.5 minutes?
b) If we assume the data follows normal distribution, approximately how many emergency calls are responded within 5 minutes?
c) If we assume the data follows a left skewed distribution, at least 68% of the emergency calls should have their response time between what range?
d) If we assume the data follows the normal distribution, for a randomly received emergency call, what is the probability that the response time will be between 6.7 to 11.8 minutes?
e) If we assume the data follow a right skewed distribution, for a randomly received emergency call, what is the probability that the response time will be between 5.34 to 11.46 minutes?
"n=5000 \\\\\n\n\u03bc=8.4 \\\\\n\n\u03c3 = 1.7 \\\\\n\n(a) \\; P(X>13.5) = P(\\frac{X-\u03bc}{\u03c3}> \\frac{13.5-8.4}{1.7}) \\\\\n\n=P(Z>3) \\\\\n\n= 1 -P(Z<3) \\\\\n\n= 1 -0.9987 \\\\\n\n= 0.0013 \\\\\n\n(b) \\; P(X\u22645) = P(\\frac{X-\u03bc}{\u03c3} \u2264 \\frac{5-8.4}{1.7}) \\\\\n\n= P(Z \u2264 -2) \\\\\n\n= 1 -P(Z\u22642) \\\\\n\n= 1 -0.4772 \\\\\n\n= 0.0228"
(c) Chebyshev’s Theorem can be applied to both left skewed and right skewed distributions.
It states that Percentage of values that lies within k standard deviations from mean = 1 - (1/k²)
"0.68 = 1- \\frac{1}{k^2} \\\\\n\n\\frac{1}{k^2} = 1- 0.68 =0.32 \\\\\n\nk = \\sqrt{\\frac{1}{0.32}} = 1.768 \\\\\n\nRange = 8.4- 1.768 \\times 1.7 = 5.3944 \\\\\n\nRange = 8.4+ 1.768 \\times 1.7=11.4056 \\\\\n\n(5.3944, 11.4056) \\\\\n\n(d) \\; P(6.7<X<11.8) = P(\\frac{6.7-8.4}{1.7} < \\frac{X-\u03bc}{\u03c3} < \\frac{11.8-8.4}{1.7}) \\\\\n\n= P(-1 <Z < 2) \\\\\n\n= P(Z<2) -P(Z <-1) \\\\\n\n= 0.9772 -0.1586 \\\\\n\n= 0.8186"
(e) Similar to the c part:
"Range = 8.4 -k \\times 1.7 = 5.34 \\\\\n\nRange = 8.4 +k \\times 1.7 = 11.46 \\\\\n\nk = 1.8 \\\\\n\nP = 1- \\frac{1}{k^2} \\\\\n\n= 1 -\\frac{1}{(1.8)^2} \\\\\n\n= 0.69"
Answer: 69 %
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