Question

An investment website can tell what devices are used to access the site. The site managers wonder whether they should enhance the facilities for trading via​ "smart phones", so they want to estimate the proportion of users who access the site that way​(even if they also use their computers​ sometimes). They draw a random sample of 100 investors from their customers. Suppose that the true proportion of smart phone users is 41%. a) What would the standard deviation of the sampling distribution of the proportion of the smart phone users​ be? b) What is the probability that the sample proportion of smart phone users is greater than 0.41? c) What is the probability that the sample proportion is between 0.19 and 0.31​?

Answer 1

a)

std.deviation = sqrt(p*(1-p)/n)
= sqrt(0.41 *(1-0.41)/100)
= 0.0492

b)

Here, μ = 0.41, σ = 0.0492 and x = 0.41. We need to compute P(X >= 0.41). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z = (0.41 - 0.41)/0.0492 = 0

Therefore,
P(X >= 0.41) = P(z <= (0.41 - 0.41)/0.0492)
= P(z >= 0)
= 1 - 0.5 = 0.5

c)

Here, μ = 0.41, σ = 0.0492, x1 = 0.19 and x2 = 0.31. We need to compute P(0.19<= X <= 0.31). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z1 = (0.19 - 0.41)/0.0492 = -4.47
z2 = (0.31 - 0.41)/0.0492 = -2.03

Therefore, we get
P(0.19 <= X <= 0.31) = P((0.31 - 0.41)/0.0492) <= z <= (0.31 - 0.41)/0.0492)
= P(-4.47 <= z <= -2.03) = P(z <= -2.03) - P(z <= -4.47)
= 0.0212 - 0
= 0.0212