Question

100 atistics om o 4 2 ary Statistics 2. In the Healthy Hndwashing Survey conducted by Bradley Corporation, it was found that 64% of adult Americans operate the flusher of toilets in public restrooms with their foot. Find probability that if 500 adult Americans are randomly selected... (a) Less than 300 of them would flush toilets in public restrooms with their foot. (b) At least 325 of them would flush toilets in public restrooms with their foot. But first, describe the binomial experiment: • Number of trials, n: Describe two possible outcomes of the experiment: • Which outcome is referred to as success? • Probability of success for each trial, p: 9. (q) (a) with 8. (Sec ..

Answer 1

Thing to describe

Number of trials n = 500

Two possible outcomes ( yes, no ) i.e adult Americans operate the flusher of toilet in public restrooms with their foot or not.

Outcome referred to success is (Yes, adult Americans do operate the flusher of toilet in public restrooms with their foot )

Probability of success P = 64% = 0.64

Part a)

Using Normal Approximation to Binomial
Mean = n * P = ( 500 * 0.64 ) = 320
Variance = n * P * Q = ( 500 * 0.64 * 0.36 ) = 115.2
Standard deviation = √(variance) = √(115.2) = 10.7331

P ( X < 300 )
Using continuity correction
P ( X < n - 0.5 ) = P ( X < 300 - 0.5 ) = P ( X < 299.5 )

X ~ N ( µ = 320 , σ = 10.7331 )
P ( X < 299.5 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 299.5 - 320 ) / 10.7331
Z = -1.91
P ( ( X - µ ) / σ ) < ( 299.5 - 320 ) / 10.7331 )
P ( X < 299.5 ) = P ( Z < -1.91 )
P ( X < 299.5 ) = 0.0281

Part b)

P ( X >= 325 )
Using continuity correction
P ( X > n - 0.5 ) = P ( X > 325 - 0.5 ) =P ( X > 324.5 )

X ~ N ( µ = 320 , σ = 10.7331 )
P ( X > 324.5 ) = 1 - P ( X < 324.5 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 324.5 - 320 ) / 10.7331
Z = 0.42
P ( ( X - µ ) / σ ) > ( 324.5 - 320 ) / 10.7331 )
P ( Z > 0.42 )
P ( X > 324.5 ) = 1 - P ( Z < 0.42 )
P ( X > 324.5 ) = 1 - 0.6628
P ( X > 324.5 ) = 0.3372