a)The probability that all 10 will arrive within two days is 0.3486.
b) P(x = 8) = 0.1937
c) P(x=5)= 0.0014
d) mean = 9
e) Standard deviation = √0.90 = 0.9486
The binomial distribution model allows us to compute the probability of observing a specified number of "successes" when the process is repeated a specific number of times and the outcome for a given patient is either a success or a failure.
Given:
p = 90/100 = 0.90
q(probability of success) = 1 - p = 1 - 0.90 = 0.10
n = 10
a)The probability that all 10 will arrive within two days
P(x = 10) = 0.3486
b) P(x = 8) = 0.1937
c) P(x=5)= 0.0014
d) mean = np = 10 × 0.90 = 9
e) Variance = npq = 10 × 0.90 × 0.10 = 0.90
Standard deviation = √0.90 = 0.9486
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