Question

D My Notes (t) Ask You o-4 points BBUnderStat12 5.3.009 I consider a binomial distribution with n = 10 trials and the probability of success on a single trial ρ = 0.95. (a) Is the distribution skewed left, skewed right, or symmetric? O skewed left O skewed right O symmetric (b) Compute the expected number of successes in 10 trials. on a single trial, would you expect P(r 4) to be very high or very (c) Given the high probability of success Explain. O Very high. The expected number of successes in 10 trials is more than 4, and p is so high that it would be unusual to have so few successes in 10 trials. y high. The expected number of successes in 10 trials is more than 4, and p is so high that it would be common to have so few successes in 10 trials. O Ve O very low. The expected number of successes in 10 trials is more than 4, and p is so high that it would be unusual to have so few successes in 10 trials. very low. The expected number of successes in 10 trials is more than 4, and p is so high that it would be common to have so few successes in 10 trials. (d) Given the high probability of success p on a single trial, would you expect Pir 2 9) to be very high or very lov Explain. O Very high. The expected number of successes in 10 trials is mare than 9, and p is so high that it would be common to have 9 or more successes in 10 trials. O Very high. The expected number of successes in 10 trials is more than 9, and p is so high that it would be o very low. The expected number of successes in 10 trials is more than 9, and ρ is so high that it would be O Very low. The expected number of successes in 10 trials is more than 9, and p is so high that it would be unusual to have 9 or more successes in 10 trials. common to have 9 or more successes in 10 trials. unusual to have 9 or more successes in 10 trials. Need Help? ReadlIt

Answer 1

n = 10 p = 0.95 p(r) = 10_c_r (0.95) ^r (0.05) ^1 rms skewed left option 2 expected number of successes 9.5 np = 10 times 0.95 = 9.5 p(r lessthanorequalto 4) = p(0) + p(1) + p(2) + p(3) + p(4) = 0.00000275 option 3 unused to have few successes in 10 trials option 1 very high, common to have 9 or more successes in 10 trails