gia9482

2022-10-16

Based on historical data, your manager believes that 30% of the company's orders come from first-time customers. A random sample of 122 orders will be used to estimate the proportion of first-time-customers. What is the probability that the sample proportion is less than 0.33?

karton

Given that the manager believes that 30% of the company's orders come from first-time customers, we can denote the true population proportion as $p$. In this case, $p=0.30$.
To find the probability that the sample proportion is less than 0.33, we need to calculate the z-score and use the standard normal distribution.
The formula for the z-score is:
$z=\frac{\stackrel{^}{p}-p}{\sqrt{\frac{p·\left(1-p\right)}{n}}}$
Where:
- $\stackrel{^}{p}$ is the sample proportion
- $p$ is the population proportion
- $n$ is the sample size
In this case, $\stackrel{^}{p}=0.33$ and $n=122$. Plugging these values into the formula, we get:
$z=\frac{0.33-0.30}{\sqrt{\frac{0.30·\left(1-0.30\right)}{122}}}$
Calculating the expression within the square root:
$\sqrt{\frac{0.30·\left(1-0.30\right)}{122}}\approx 0.036$
Substituting this value into the formula for z:
$z=\frac{0.33-0.30}{0.036}\approx 0.833$
Now, we can find the probability using the standard normal distribution table or a statistical calculator. The probability that the sample proportion is less than 0.33 corresponds to the area to the left of the z-score of 0.833.
$P\left(\stackrel{^}{p}<0.33\right)=P\left(z<0.833\right)$
Using the standard normal distribution table or calculator, we find that this probability is approximately 0.7967.
Therefore, the probability that the sample proportion is less than 0.33 is approximately 0.7967 or 79.67%.

Do you have a similar question?