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In humans the intermidiate stage of sleep is characterized by the presence of highamplitude waves averaging about 2 waves per second. Find the probability of observing 15 or more highamplitude waves in a fivesecond period of intermediate sleep. Please help I'm pretty lost for some reason and I have to turn this in at 12:30
 one year ago
 one year ago
In humans the intermidiate stage of sleep is characterized by the presence of highamplitude waves averaging about 2 waves per second. Find the probability of observing 15 or more highamplitude waves in a fivesecond period of intermediate sleep. Please help I'm pretty lost for some reason and I have to turn this in at 12:30
 one year ago
 one year ago

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CarlosGPBest ResponseYou've already chosen the best response.0
Too late for your assignment but on time to learn. Whenever you have a problem related to rates (such as the case of waves per second or calls per minute in a call center or fires per hour for a fire department and so on...) you need to apply Poisson Distribution. The formula is: \[P(X=k)=\lambda^ke^{\lambda}/k!\]Lambda (λ) is the number of occurrences during the period of observation. In our case λ=(2 waves/second) x (5 seconds)=10 waves in 5 seconds as average rate. Your formula will be then: \[P(X=k)=10^ke^{10}/k!\] The probability of observing less than 15 waves is the probability of observing up to 14 waves (in other words observing 0,1,2,3....14 waves). This can be expressed as:\[P(X < 15)=P(X \le 14)=\sum_{k=0}^{k=14}P(X=k)\] and the probability of observing 15 or more is\[P(X \ge 15)=1P(X<15)=1P(X \le 14)\] All this can be put as: \[P(X \ge 15)=1\sum_{k=0}^{k=14}10^ke^{10}/k!\] If you use Excel to calculate it, you get:\[P(X \ge 15)=10.9165=0.0835\] and that is the same than 8.35%
 one year ago
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