Compute the following binomial probabilities directly from the formula for b(x,n,p): a) b(3, 8, 0.6) b) b(5, 8, 0.6) c) P(3≤X≤5) when n=8 and p=0.6 d)P(1≤X) when n=12 and p=0.1

Question

Compute the following binomial probabilities directly from the formula for b(x,n,p):  a) b(3, 8, 0.6)  b) b(5, 8, 0.6)  c) P(3≤X≤5)  when n=8 and p=0.6  d)P(1≤X) when n=12 and p=0.1

comentezq · Accepted Answer

Step 1a) Given:n=8,p=0.6,x=3X∼&amp;nbsp;Binomeal&amp;nbsp;(x=8,p=0.6)&amp;nbsp;I will use the excel formula &quot;=BINOM.DIST(x, n, p, FALSE)&quot; to find the probability.&amp;nbsp;xnpExcelformulaProbability,P(x)Roundtothreedecimals380.6=BINOM.DIST(3,8,0.6,FALSE)0.124As a result, this probability is&amp;nbsp;P(3)=0.124Answer: 0.124Step 2b) Given:&amp;nbsp;n=8,p=0.6,x=5&amp;nbsp;X∼Binomial&amp;nbsp;(x=8,p=0.6)&amp;nbsp;I intend to use the Excel formula &quot;=BINOM.DIST(x, n, p, FALSE)&quot; to find the probability.xnpExcelformulaProbability,P(x)Roundtothreedecimals580.6=BINOM.DIST(5,8,0.6,FALSE)0.279Therefore, this probability is&amp;nbsp;P(5)=0.279Answer: 0.279Step 3c) Given:&amp;nbsp;n=8,p=0.6X∼&amp;nbsp;Binomial&amp;nbsp;(n=8,p=0.6)&amp;nbsp;find:&amp;nbsp;P(3≤X≤5)=P(3)+P(4)+P(5)&amp;nbsp;I intend to use the Excel formula &quot;=BINOM.DIST(x, n, p, FALSE)&quot; to find the probability.&amp;nbsp;xnpExcelformulaProbability,P(x)Round&amp;nbsp;to&amp;nbsp;three&amp;nbsp;decimals380.6=BINOM.DIST(3,8,0.6,FALSE)0.124480.6=BINOM.DIST(4,8,0.6,FALSE)0.232580.6=BINOM.DIST(5,8,0.6,FALSE)0.279Therefore, this probability is&amp;nbsp;P(3≤X≤5)=0.124+0.232+0.279P(3≤X≤5)=0.635Answer: 0.635Step 4d) Given:&amp;nbsp;n=12,p=0.1X∼&amp;nbsp;Binomial&amp;nbsp;(n=12,p=0.1)&amp;nbsp;find:&amp;nbsp;P(1≤X)=?&amp;nbsp;Use the complement rule,&amp;nbsp;P(1≤X)=1−P(0)&amp;nbsp;I will make use of the Excel formula &quot;=BINOM.DIST(x, n, p, FALSE)&quot; to find the probability.

nick1337 · Accepted Answer

Step 1:a) To solve the binomial probability b(3,8,0.6), we can use the formula:b(x,n,p)=(nx)·px·(1−p)n−xSubstituting the given values, we have:b(3,8,0.6)=(83)·(0.6)3·(1−0.6)8−3Calculating the binomial coefficient:(83)=8!3!(8−3)!=8!3!·5!=8·7·63·2·1=56Substituting the values:b(3,8,0.6)=56·(0.6)3·(1−0.6)8−3Evaluating the expression:b(3,8,0.6)=56·0.63·0.45≈0.2784Therefore, b(3,8,0.6)≈0.2784.Step 2:b) To compute the binomial probability b(5,8,0.6), we use the same formula:b(5,8,0.6)=(85)·(0.6)5·(1−0.6)8−5Calculating the binomial coefficient:(85)=8!5!(8−5)!=8!5!·3!=8·7·63·2·1=56Substituting the values:b(5,8,0.6)=56·(0.6)5·(1−0.6)8−5Evaluating the expression:b(5,8,0.6)=56·0.65·0.43≈0.258048Therefore, b(5,8,0.6)≈0.258048.Step 3:c) To compute the probability P(3≤X≤5) when n=8 and p=0.6, we sum the individual binomial probabilities for x=3,4, and 5:P(3≤X≤5)=b(3,8,0.6)+b(4,8,0.6)+b(5,8,0.6)Substituting the values:P(3≤X≤5)=b(3,8,0.6)+b(4,8,0.6)+b(5,8,0.6)Using the values calculated in parts (a) and (b):5)≈0.2784+0.258048+0.258048Simplifying the expression:P(3≤X≤5)≈0.794496Therefore, P(3≤X≤5)≈0.794496.Step 4:d) To compute the probability P(1≤X) when n=12 and p=0.1, we need to sum the individual binomial probabilities for all values of x starting from 1 up to n=12:P(1≤X)=b(1,12,0.1)+b(2,12,0.1)+…+b(12,12,0.1)Since we need to calculate a large number of terms, it&#039;s more convenient to calculate the complementary probability P(X=0) and subtract it from 1:P(1≤X)=1−P(X=0)To calculate P(X=0), we use the formula:P(X=0)=(120)·(0.1)0·(1−0.1)12−0=(1−0.1)12Simplifying:P(X=0)=(0.9)12Substituting in the complementary probability formula:P(1≤X)=1−(0.9)12Calculating the expression:P(1≤X)≈1−0.28243Therefore, P(1≤X)≈0.71757.

Don Sumner · Accepted Answer

a) To compute the binomial probability b(x,n,p) for x=3, n=8, and p=0.6, we use the formula:b(3,8,0.6)=(83)·(0.6)3·(1−0.6)8−3Let&#039;s calculate each component of the formula:(83)=8!3!(8−3)!=8·7·63·2·1=56Therefore, b(3,8,0.6)=56·(0.6)3·(0.4)5.b) To compute the binomial probability b(x,n,p) for x=5, n=8, and p=0.6, we use the same formula:b(5,8,0.6)=(85)·(0.6)5·(1−0.6)8−5Calculating each component:(85)=8!5!(8−5)!=8·7·63·2·1=56Therefore, b(5,8,0.6)=56·(0.6)5·(0.4)3.c) To compute the probability P(3≤X≤5) when n=8 and p=0.6, we sum the individual probabilities for x=3, 4, and 5:P(3≤X≤5)=b(3,8,0.6)+b(4,8,0.6)+b(5,8,0.6)Substituting the values we calculated earlier:P(3≤X≤5)=56·(0.6)3·(0.4)5+56·(0.6)4·(0.4)4+56·(0.6)5·(0.4)3d) To compute the probability P(1≤X) when n=12 and p=0.1, we need to sum the individual probabilities for all values of x from 1 to 12:P(1≤X)=b(1,12,0.1)+b(2,12,0.1)+…+b(12,12,0.1)We can simplify this calculation using the complement rule. Since P(1≤X)=1−P(X=0), we can find the probability of X=0 and subtract it from 1:P(X=0)=b(0,12,0.1)=(120)·(0.1)0·(1−0.1)12−0Calculating each component:(120)=1Therefore,P(X=0)=(120)·(0.1)0·(1−0.1)12−0Calculating each component:(120)=1Therefore, P(X=0)=1·(0.1)0·(0.9)12.Now, we can calculate P(1≤X) using the complement rule:P(1≤X)=1−P(X=0)=1−(0.9)12Thus, the binomial probability P(1≤X) when n=12 and p=0.1 is 1−(0.9)12.

Compute the following binomial probabilities directly from the formula for b(x, n, p):a) b(3, 8, .6)b) b(5, 8, .6)c) P(3 ≤ X ≤ 5) when n = 8 and p = .6d)P(1 ≤ X) when n = 12 and p = .1

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