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8. If $X \sim Normal(45, 15)$, determine each unknown value (some are probabilities, some are values of X). (a) $P(X \le 55) = ?$ (b) $P(X \ge 62) = ?$ (c) $P(30 \le X \le 40) = ?$ (d) $P(|X - 42| < 5) = ?$ (e) $P(X \le ?) = 0.15$ (f) $P(X \ge ?) = 0.60$ (g) $P(50 \le X \le ?) = 0.22$ (h) $P(? \le X \le 60) = 0.60$ 

          8. If $X \sim Normal(45, 15)$, determine each unknown value (some are probabilities, some are values of X).
(a) $P(X \le 55) = ?$
(b) $P(X \ge 62) = ?$
(c) $P(30 \le X \le 40) = ?$
(d) $P(|X - 42| < 5) = ?$
(e) $P(X \le ?) = 0.15$
(f) $P(X \ge ?) = 0.60$
(g) $P(50 \le X \le ?) = 0.22$
(h) $P(? \le X \le 60) = 0.60$
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8. If X ∼ Normal(45, 15), determine each unknown value (some are probabilities, some are values of X). (a) P(X ≤ 55) = ? (b) P(X ≥ 62) = ? (c) P(30 ≤ X ≤ 40) = ? (d) P(|X - 42| < 5) = ? (e) P(X ≤ ?) = 0.15 (f) P(X ≥ ?) = 0.60 (g) P(50 ≤ X ≤ ?) = 0.22 (h) P(? ≤ X ≤ 60) = 0.60

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Elementary Statistics a Step by Step Approach
Elementary Statistics a Step by Step Approach
Allan G. Bluman 9th Edition
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I need help answering d-h. If someone could please show the steps on how to solve them! Thank you :) 8. If X ~ Normal(45,15), determine each unknown value. Some are probabilities, some are values of X. a) P(X < 55) = b) P(X > 62) = c) P(30 < X < 40) = d) P(1 < X - 42 < 5) = e) P(X < 3) = 0.15 f) P(X = 0.60) = g) P(50 < X) = 0.22 h) P(X > 60) = 0.60
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Transcript

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0:00 All right.
00:01 On the given question, we have been asked to determine the probability for the given normal distribution problems.
00:07 So for the first part, they say that the population mean is 50.
00:12 Standard deviation is 34 .1.
00:14 And we have been asked to determine the probability that x is less than equal to 43.
00:20 Now in the z score formula, we know this is x minus mean upon standard deviation.
00:25 So this will be equal to probability of x minus mean upon standard deviation less than equal to 43 mean here is 50 divided by it is 34 .1 so this will be equal to probability of z less than minus 0 .21 and this you can determine from the z table this is value 0 .468 right now next part of the question for the next here it says that the population mean is one zero two and standard deviation is 24 and we have to determine the probability that x is greater than 84 so this will be equal to the same formula we'll be using here again that is x minus mean upon standard deviation is greater than 84 minus 1002 upon 24 so this will be equal to probability of z greater than minus 0 .75 right so this will be equal to one minus probability of z greater than minus 0 .75 right so this will be equal to 1 minus probability of z less than minus 0 .75 so this will be equal to 0 .7734 right moving on to the next part here okay so here we have the population mean is given to us as 72 and standard deviation is 13 and we have to determine the probability that x is uh in between wait okay it is between 55 and 81 right so here this will be equal to 55 minus 72 divided by 13 less than equal to x minus mean upon standard deviation less than equal to 81 minus 72 divided by 13 right so that is equal to okay so this will be minus 1 .31 less than z less than 0 .69 so this will be equal to probability of z less than 0 .69 minus probability of z less than 0 .75 or sorry this is this is minus 1 .31 right so this will be equal to 0 .7549 minus 0 .099 that is equal to 0 .0659 that is equal to 0 .06598 right and for the last part we have the population mean as 11 standard deviation is 3 .1 so 3 .4 and we have to determine the probability that x lies in between 12 and 16 so this will be probability 12 minus 11 upon 3 .4 less than x minus mean upon standard deviation 16 minus 11 upon 3 .4 right so this is equal to probability of 0 .29 less than z less than 1 .47...
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