A local newspaper claims that 90% of its online readers are under the age

Question

Jazz Frenia · Accepted Answer

To solve this problem, we will use the sample proportion and the population proportion to calculate the probability.
Let p be the population proportion of online readers under the age of 45 years, which is claimed to be 90%. Therefore, p = 0.9.
We are given a sample of 300 online readers, where 240 are under the age of 45 years. The sample proportion, denoted as p-hat (p̂), is calculated by dividing the number of readers under the age of 45 (240) by the sample size (300). Therefore, p̂ = 240/300 = 0.8.
To find the probability that the sample proportion is more than 85%, we need to calculate the z-score and use the standard normal distribution.
The formula for calculating the z-score for a sample proportion is:

z = \frac{p ̂ - p}{\sqrt{\frac{p (1 - p)}{n}}}

Substituting the values, we have:

z = \frac{0.8 - 0.9}{\sqrt{\frac{0.9 (1 - 0.9)}{300}}}

Calculating this expression:

z = \frac{- 0.1}{\sqrt{\frac{0.9 (0.1)}{300}}}

z = \frac{- 0.1}{\sqrt{\frac{0.09}{300}}}

z = \frac{- 0.1}{\sqrt{0.0003}}

z = \frac{- 0.1}{0.0173205}

z \approx - 5.7735

Now, we need to find the probability that the z-score is greater than -5.7735. This can be calculated using a standard normal distribution table or a calculator.

P (Z > - 5.7735)

is almost equal to 1.
Therefore, the probability that the sample proportion of the online readers under the age of 45 years is more than 85% is approximately 1, or 100%.

A local newspaper claims that 90% of its

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