Question

The sampling distribution of the percentage of adults between the ages of 30 and 49 years who say they use Facebook in many samples of size n = 520, will be Normal in shape, with mean (or center) of 78% (or 0.78) and standard deviation of 1.8% (or 0.018).   Based on this information, we know that the probability of obtaining a sample of size n = 520 adults between the ages of 30 and 49 years where 75% or fewer say they use Facebook is ...

1. 0.0227 2. 0.0359 3. 0.0446 4. 0.0500 5. 0.1357

What is the probability of obtaining a sample of size n = 520 adults between the ages of 30 and 49 years where 80% or more say they use Facebook?

1. 0.1357 2. 0.0500 3. 0.1151 4. 0.8643 5. 0.9332

Answer 1

= 0.78

= 1.8/100 = 0.018

X ~ Normal(0.78, 0.018)

P(75% or fewer ) = P(X 0.75) = P[Z (0.75 - 0.78)/0.018]

= P[Z -1.7]

= 0.0446

P(80% or more) = P(X 0.80) = P[Z (0.80 - 0.78)/0.018]

= P[Z 1.1]

= 0.1357