Let the distribution of X be Binomial(150,0.8). The next 4 questions correspond to this information. The answer may be rounded up to 3 decimal places of the actual value:

Now the expression " P(X < 121)" does not include 121 so to calculate this interval you have to substract the cummulative probability to the previoous value of the variable " P(X ≤ 120)"

Then the interval is

P(X ≤ 129) - P(X ≤ 120)= 0.978 - 0.533= 0.445

2) The approximation to normal (without correction of continuity) of P(121 < X ≤ 129) is:

To use the normal approximation you have to calculate the mean and variance of the variable.

E(X)= np= 150*0.8= 120

V(X)= np(1-p)= 150*0.8*0.2= 24

Now you can standardize the given interval:

P(X ≤ 129) - P(X < 121)= P(Z ≤ (129-120)/√24) - P(Z < (121-120)/√24)

P(Z ≤ 1.84) - P(Z < 0.20) = 0.967 - 0.579= 0.388

3) The approximation to normal (with correction of continuity) of P(121 < X ≤ 129) is:

P(121 < X ≤ 129)

Applying the correction of continuity:

For X ≤ n + 0.5

For X > n + 0.5

P(121.5 < X ≤ 129.5) = P(X ≤ 129.5) - P(X < 121.5)

P(Z ≤ (129.5-120)/√24) - P(Z < (121.5-120)/√24)

P(Z ≤ 1.94) - P(Z < 0.31) = 0.974 - 0.622= 0.352

4) The approximation of poisson of P(121 < X ≤ 129) is:

First define the rate of successes of the distribution λ= np= 150*0.8= 120

Then you look at the individual cummulative probabilities using the tables of the distribution:

P(X ≤ 129; λ= 120)= 0.808

P(X < 121; λ= 120)= P(X ≤ 120; λ= 120)= 0.524

P(121 < X ≤ 129) = P(120 ≤ X ≤ 129)= P(X ≤ 129) - ≤ 129)= 0.808 - 0.524= 0.284

I hope it helps!