Question

The Town of Hertfordshire clerk knows that 23% of dogs in the town have completed emotional support training. Hertfordshire plans on showcasing a simple random sample of its dogs in a show. Depending on which dogs are chosen, the proportion of emotional support trained dogs may vary. In a sample of 100 dogs, what is the probability that under 30% of the dogs are emotional support trained?

Answer 1

Answer:

95.64% probability that under 30% of the dogs are emotional support trained

Step-by-step explanation:

For each dog, there are only two possible outcomes. Either they have completed emotional support training, or they have not. So we use the binomial probability distribution to solve this problem.

However, we are working with samples that are considerably big. So i am going to aproximate this binomial distribution to the normal.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

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

The Z-score 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. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$, $\sigma = \sqrt{V(X)}$.

In this problem, we have that:

$\mu = E(X) = np = 100*0.23 = 23$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{100*0.23*0.77} = 4.21$

What is the probability that under 30% of the dogs are emotional support trained?

30% of 100 is 0.3*100 = 30

So this is the pvalue of Z when X = 30.

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

$Z = \frac{30 - 23}{4.1}$

$Z = 1.71$

$Z = 1.71$ has a pvalue of 0.9564.

So there is a 95.64% probability that under 30% of the dogs are emotional support trained