<span>The solution:
= 40, p = q = 0.5
P[x] = nCx *p^x *q^(n-x)
when p = q = 0.5, the formula simplifies to
P[x] = nCx/2^n = 40Cx/2^40
at least 18 of each type means 18 to 22 of (say) type I
P(18 <= X <= 22) = 0.5704095 <-------
qb
mean = 40*0.5 = 20
SD = sqrt(npq) = sqrt(40*0.5*0.5) = 3.1623
z1= (18-20)/3.1623 = -0.63 , z2 = (22-20)/3.1623 = 0.63
P(-0.63 < z < 0.63) = 0.4713 <-------</span>
C is the only one greater than 12.5 i think so
also b and d
2.6 rounds to 3, and 0.33 would round to 0.50. Adding 3 and 0.50 together would equal to 3.50. 59 and 3.50 are no where near to each other which concludes Minh is incorrect.
(Please mark me brainliest)
Try explaining how you worked the problem and how you did it
