Question

During a drug trail for a new antibiotic, 45% of the people who were given the drug experienced dizziness. Assume a sample size of 15 patients..

a) FInd the probability that exactly 8 patients experienced dizziness.

B)FInd the probability that less than 6 patients experienced dizziness.

C)FInd the probability that at least 7 patients experienced dizziness.

D)Find the probability that at most 4 patients did NOT experience dizziness..

Answer 1

During a drug trail for a new antibiotic, 45% of the people who were given the drug experienced dizziness. Assume a sample size of 15 patients..

a) Find the probability that exactly 8 patients experienced dizziness.

B) Find the probability that less than 6 patients experienced dizziness.

C) Find the probability that at least 7 patients experienced dizziness.

D) Find the probability that at most 4 patients did NOT experience dizziness..

Solution

A) Find the probability that exactly 8 patients experienced dizziness.

X ~ Binomial(15, 0.45)

$$P (X = 8) = \binom {1 5} {8} (0. 4 5) ^ {8} (0. 5 5) ^ {1 5 - 8} = 0. 1 6 4 7 4$$

B) Find the probability that less than 6 patients experienced dizziness.

X ~ Binomial(15, 0.45)

$$P (X < 6) = \sum_ {x = 0} ^ {5} \binom {1 5} {x} (0. 4 5) ^ {x} (0. 5 5) ^ {1 5 - x} = 0. 2 6 0 7 6$$

C) Find the probability that at least 7 patients experienced dizziness.

X ~ Binomial(15, 0.45)

$$\mathrm {P} (\mathrm {X} \geq 7) = \sum_ {x = 7} ^ {1 5} \binom {1 5} {x} (0. 4 5) ^ {x} (0. 5 5) ^ {1 5 - x} = 0. 5 4 7 8 4$$

D) Find the probability that at most 4 patients did NOT experience dizziness.

X ~ Binomial(15, 0.55)

$$\mathrm {P} (\mathrm {X} \leq 4) = \sum_ {x = 0} ^ {4} \binom {1 5} {x} (0. 5 5) ^ {x} (0. 4 5) ^ {1 5 - x} = 0. 0 2 5 4 7$$

Answer: A) 0.16474; B) 0.26076; C) 0.54784; D) 0.02547.