Question

2. A certain antibiotic is known to cure 85% of streptococcal bacterial infections. A scientist wants to make sure that the drug does not lose its potency over time. He treats 100 streptococcal patients with a one year old supply of the antibiotic.

(a) What are the mean and standard deviation of the sampling distribution of the sample proportion of individuals who are cured? State any required assumptions.

(b) What is the probability that more than 90% in the sample are cured?

Answer 1

Solution

Given that,

p = 0.85

1 - p = 1-0.85=0.15

n = 100

(A) mean = = p =0.85

standard deviation= =  [p( 1 - p ) / n] = [(0.85*0.15) /100 ] = 0.0357

(B)P( > 0.90) = 1 - P( <0.90 )

= 1 - P(( - ) / < (0.90-0.85) / 0.0357)

= 1 - P(z <1.40 )

Using z table

= 1 -0.9192

=0.0808

probability=0.0808