Question

question 5
it is known that 45% of customers will need additional maintenance on their vehicle when coming in for an oil change. a random sample of 48 customers is taken. normal distribution.
what does the standard error equal?
what is the probability that less than 53% of customers in the sample will need additional maintenance?

Explanation:

Step1: Calculate the standard error

The formula for the standard error of a proportion is $SE = \sqrt{\frac{p(1 - p)}{n}}$, where $p = 0.45$ (population proportion) and $n = 48$ (sample size).
$$SE=\sqrt{\frac{0.45\times(1 - 0.45)}{48}}=\sqrt{\frac{0.45\times0.55}{48}}=\sqrt{\frac{0.2475}{48}}\approx\sqrt{0.00515625}\approx0.0718$$

Step2: Calculate the z - score

The z - score formula is $z=\frac{\hat{p}-p}{SE}$, where $\hat{p}=0.53$ (sample proportion), $p = 0.45$, and $SE\approx0.0718$.
$$z=\frac{0.53 - 0.45}{0.0718}=\frac{0.08}{0.0718}\approx1.11$$

Step3: Find the probability

We want to find $P(\hat{p}<0.53)$, which is equivalent to $P(Z < 1.11)$ using the standard normal distribution. Looking up the value in the standard normal table or using a calculator, $P(Z < 1.11)=0.8665$

Answer:

The standard error is approximately $0.0718$ and the probability that less than $53\%$ of customers in the sample will need additional maintenance is approximately $0.8665$