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Answer:
(a) The probability that a single randomly selected value lies between 158.6 and 159.2 is 0.004.
(b) The probability that a sample mean is between 158.6 and 159.2 is 0.0411.
Step-by-step explanation:
Let the random variable <em>X</em> follow a Normal distribution with parameters <em>μ</em> = 155.4 and <em>σ</em> = 49.5.
(a)
Compute the probability that a single randomly selected value lies between 158.6 and 159.2 as follows:
*Use a standard normal table.
Thus, the probability that a single randomly selected value lies between 158.6 and 159.2 is 0.004.
(b)
A sample of <em>n</em> = 246 is selected.
Compute the probability that a sample mean is between 158.6 and 159.2 as follows:
*Use a standard normal table.
Thus, the probability that a sample mean is between 158.6 and 159.2 is 0.0411.
Cos (B) = (a^2 + c^2 -b^2) / (2 * a * c)
cos (B) = (11^2 +17^2 -12^2) / (2 * 11 * 17)
cos (B) = (121 + 289 -144) / (374)
cos (B) = 266 / 374
cos (B) = <span> <span> <span> 0.7112299465
Angle B = </span></span></span>44.665 degrees
cos (B) = (11^2 +17^2 -12^2) / (2 * 11 * 17)
cos (B) = (121 + 289 -144) / (374)
cos (B) = 266 / 374
cos (B) = <span> <span> <span> 0.7112299465
Angle B = </span></span></span>44.665 degrees