A bank manager wishes to provide prompt service for customers at the bank's drive-up window. The bank currently can serve up to 10customers per 15-minute period without significant delay. The average arrival rate is 7 customers per 15-minute period. Let x denotethe number of customers arriving per 15-minute period. Assuming x has a Poisson distribution:(a) Find the probability that 10 customers will arrive in a particular 15-minute period. (Round your answer to 4 decimal places.)Probability(b) Find the probability that 10 or fewer customers will arrive in a particular 15-minute period. (Do not round intermediate calculationsfor calculating probability. Round your answer to 4 decimal places.)Probability(c) Find the probability that there will be a significant delay at the drive-up window. That is, find the probability that more than 10customers will arrive during a particular 15-minute period. (Do not round intermediate calculations for calculating probability. Roundyour answer to 4 decimal places.)Probability

GivenA random variable x ~poisson(λ)P(X=x)=e-λ×λxx! ; x≥0a)The probability that 10 customers will arrive in a particular 15-minute period isλ=7 per 15-minute periodRequired probability P(x=10)=e-7×71010!=0.0710Probability is 0.0710b)The probability that 10 or fewer cutomers will arrive in a particular 15-minute period is=P(x≤10)=P(x=0)+P(x=1)+P(x=2)+P(x=3)+P(x=4)+P(x=5)+P(x=6)+P(x=7)+P(x=8)+P(x=9)+P(x=10)=e-7×700!+e-7×711!+e-7×722!+e-7×733!+e-7×744!+e-7×755!+e-7×766!+e-7×777!+e-7×788!+e-7×799!+e-7×71010!=0.00091+0.00638+0.02234+0.05213+0.09123+0.12772+0.149+0.149+0.13038+0.1014+0.07098=0.9015probability is 0.9015c)The probability that there will be a significant delay at the drive-up windowλ=10 per 15-min period without significant delay=P(x>10)=1-P(x≤10)=1-[P(x=0)+P(x=1)+P(x=2)+P(x=3)+P(x=4)+P(x=5)+P(x=6)+P(x=7)+P(x=8)+P(x=9)+P(x=10)]=1-[e-10×1000!+e-10×1011!+e-10×1022!+e-10×1033!+e-10×1044!+e-10×1055!+e-10×1066!+e-10×1077!+e-10×1088!+e-10×1099!+e-10×101010!]=1-[0.5830]=0.4170probability is 0.4170