solution:
The probability mass function for binomial distribution is,
Where,
X=0,1,2,3,…..; q=1-p
find the probability that (p∧ ≤ 0.06) , substitute the values of sample units (n) , and the probability of nonconformities (p) in the probability mass function of binomial distribution.
Consider x to be the number of non-conformities. It follows a binomial distribution with n being 50 and p being 0.03. That is,
binomial (50,0.02)
Also, the estimate of the true probability is,
p∧ = x/50
The probability mass function for binomial distribution is,
Where,
X=0,1,2,3,…..; q=1-p
The calculation is obtained as
P(p^ ≤ 0.06) = p(x/20 ≤ 0.06)
= 50cx ₓ (0.03)x ₓ (1-0.03)50-x
= (50c0 ₓ (0.03)0 ₓ (1-0.03)50-0 + 50c1(0.03)1 ₓ (1-0.03)50-1 + 50c2 ₓ (0.03)2 ₓ (1-0.03)50-2 +50c3 ₓ (0.03)3 ₓ (1- 0.03)50-3 )
=( ₓ (0.03)0 ₓ (1-0.03)50-0 + ₓ (0.03)1 ₓ (1-0.03)50-1 + ₓ (0.03)2 ₓ (1-0.03)50-2 ₓ (0.03)3 ₓ (1-0.03)50-3 )
The answer to this question is choice option letter C
The Given Sequence is an Arithmetic Sequence with First term = -19
⇒ a = -19
Second term is -13
We know that Common difference is Difference of second term and first term.
⇒ Common Difference (d) = -13 + 19 = 6
We know that Sum of n terms is given by : 
Given n = 63 and we found a = -19 and d = 6
The Sum of First 63 terms is 10521
