Question

Suppose that 71% of all adults think that the overall safety of airline travel is either good or excellent. An opinion poll plans to ask an SRS of 1097 adults about airplane safety. The proportion of the sample who think that the overall safety of airline travel is either good or excellent will vary if we take many samples from this same population. The sampling distribution of the sample proportion is approximately Normal with mean 0.71 and standard deviation about 0.012. Sketch this Normal curve and use it to answer the following questions. (Round your answers to three decimal places.)

(a) What is the probability that the poll gets a sample in which more than 73.4% of the people think that the overall safety of airline travel is either good or excellent?

(b) What is the probability of getting a sample that misses the truth (71%) by 2.4% or more?

Answer 1

Solution:

Let X represents the sampling distribution of sample proportion.

X ~ N(0.71, 0.012²)

i.e. mean (μ) = 0.71 and standard deviation (σ) = 0.012

a) We have to obtain P(X > 73.4%).

73.4% = 0.734

We know that if X ~ N(μ, σ^{​​​​​​2}) then,

P(X > 73.4%) = P(X > 0.734)

Using "NORM.S.DIST" function of excel we get,

P(Z > 2) = 0.023

The probability that the poll gets a sample in which more than 73.4% of the people think that the overall safety of airline travel is either good or excellent is 0.023.

b) If sample misses the truth (71%) by 2.4% or more, then it means that either the sample proportion is 2.4% or more less than the truth or sample proportion is 2.4% or more greater than the truth.

2.4% = 0.024 and 71% = 0.71

0.71 - 0.024 = 0.686 and 0.71 + 0.024 = 0.734

Hence, we have to obtain P(X ≤ 0.686 or X ≥ 0.734).

P(X ≤ 0.686 or X ≥ 0.734) = P(X ≤ 0.686) + P(X ≥ 0.734)

We know that if X ~ N(μ, σ^{​​​​​​2}) then,

Using "NORM.S.DIST" function of excel we get,

P(Z ≤ -2) = 0.023 and P(Z ≥ 2) = 0.023

Hence, P(X ≤ 0.686 or X ≥ 0.734) = 0.046

The probability of getting a sample that misses the truth (71%) by 2.4% or more is 0.046.