Question

In the healthy handwashing survey, it was found the 64% of adult Americans operate the flusher of toilets in public restrooms with their foot. Suppose you survey a random sample of 740 adult American women aged 18-24 years. Use normal approximation to the binomial to approximate the probability of following.

a) check the conditions for normal distribution approximation

c)determine probability of exactly 500 of those surveyed flush toilets in public restrooms with their foot.

d) determine probability of no more than 490 of those surveyed flush toilets in public restrooms with their foot.

Answer 1

P[ adult Americans operate the flusher of toilets in public restrooms with their foot ] = p = 64% = 0.64

sample size = 740

a) check the conditions for normal distribution approximation:

a not too large value for p and q

The large value of n

Both conditions are met.

Mean = n*p = 740*0.64 = 473.6

V(x) = n*p*(1-p) = 740*0.64*(1-0.64) = 740*0.64*0.36 = 170.496

sd = sqrt(V(x)) = sqrt(170.496) = 13.0574

c)determine the probability of exactly 500 of those surveyed flush toilets in public restrooms with their foot.

P[ X = 500 ]

Normal is an continous distribution so P[ x = k ] = 0

P[ X = 500 ] = 0

d) determine the probability of no more than 490 of those surveyed flush toilets in public restrooms with their foot

P[ X < 490 ]

We need to compute . The corresponding z-value needed to be computed:

Therefore,