Question

A local agricultural cooperative claims that 55% of about 60,000 adults in a country believe that gardening should be part of the school curriculum. What is the probability that, from a simple random sample of 300 adults in the county, less than 50% would say they believe that gardening should be part of the school curriculum

Answer 1

Using the normal distribution and the central limit theorem, it is found that there is a 0.0409 = 4.09% probability that, from a simple random sample of 300 adults in the county, less than 50% would say they believe that gardening should be part of the school curriculum.

In a normal distribution with mean $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

• It measures how many standard deviations the measure is from the mean.  

• After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.

• By the Central Limit Theorem, the sampling distribution of sample proportions for a proportion p in a sample of size n has $\mu = p, s = \sqrt{\frac{p(1 - p)}{n}}$

In this problem:

• The proportion is of 55%, hence $p = 0.55$

• The sample has 300 adults, hence $n = 300$

Then, the mean and the standard error are given by:

$\mu = p = 0.55$

$s = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.55(0.45)}{300}} = 0.0287$

The probability is the p-value of Z when X = 0.5, hence:

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{0.5 - 0.55}{0.0287}$

$Z = -1.74$

$Z = -1.74$ has a p-value of 0.0409.

0.0409 = 4.09% probability that, from a simple random sample of 300 adults in the county, less than 50% would say they believe that gardening should be part of the school curriculum.

A similar problem is given at brainly.com/question/25800303