Binomial Distribution Math help?

this is a binomdist with n = 50, p = 0.52, q = 1-p = 0.48,

formula P[k] = 50Ck *0.52^k *0.48^(50-k)

a. P[≤15]=P[0] +P[1] + .... +P[15] using above formula = .00136 <------

c. P[>15] = 1 - P[≤15] = 1 - .00136 = .99864 <-------

b. P[≥15]=P[>15]+P[15]=.99864+ 50c15*.52^15*.48^35 = .9995 <-----

note that i've changed the order of answers to reduce work and also illustrate one calculation using the formula

a million.) n :- style of trials p :- danger of fulfillment on a given trial q = a million-p :- danger of failure on a given trial undergo in ideas, for a sufficiently super pattern length, a binomial distribution could be approximated with the aid of a classic distribution. we are in a position to apply here rule: If np > 5 and nq > 5, then the pattern length is adequately super, and we are in a position to approximate with the aid of a classic distribution. So calculate np and nq in each and every case, and spot whether or no longer they are extra advantageous than 5. 2.) First attempt whether np > 5 and nq > 5. If this holds, then a binomial distribution could be approximated with the aid of a classic distribution: N(pq, npq) or N(pq, np(a million-p) ) the place np is the mean (?) and npq is the variance (?^2) of the conventional distribution. To calculate a z-score, use the formula: z = (x - ?)/? Plug on your values for x (given in the question), ? = pq, and ? = squareroot(npq) to get your answer. 3.) There are sixty 5 questions, so the style of trials n = sixty 5 Mike has a a million in 5 danger of answering each and every question wisely, so p = a million/5 = 0.2 x = 10 questions we could desire to locate the risk of answering under 10 questions wisely: P(X<10) you will could desire to transform x = 10 right into a z-score making use of a similar formula as earlier: z = (x - ?)/? the place ? = pq, and ? = squareroot(npq) in case you probably did it wisely, you may get: z = -0.93026 Then making use of your z-tables, locate P(Z<z) = P(Z<-0.ninety 3)