Question

determine whether the finite correction factor should be used. if so, use it in your calculations when you find the probability. in a sample of 100 eruptions of the old faithful geyser at yellowstone national park, the mean interval between eruptions was 129.57 minutes and the standard deviation was 108.54 minutes. a random sample of size 35 is selected from this population. what is the probability that the mean interval between eruptions is between 120 minutes and 140 minutes? click here to view information about the finite correction factor. the probability that the mean interval between eruptions is between 120 minutes and 140 minutes is . (round to four decimal places as needed.)

Explanation:

Step1: Check finite - correction factor condition

The finite - correction factor is used when $n/N\geq0.05$, where $n$ is the sample size and $N$ is the population size. Here, assume the population size $N = 100$ and sample size $n=35$. Since $\frac{n}{N}=\frac{35}{100}=0.35\geq0.05$, we use the finite - correction factor. The standard deviation of the sample mean with the finite - correction factor is $\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}\sqrt{\frac{N - n}{N - 1}}$, where $\sigma = 108.54$, $n = 35$, and $N = 100$.
$\sigma_{\bar{x}}=\frac{108.54}{\sqrt{35}}\sqrt{\frac{100 - 35}{100 - 1}}\approx\frac{108.54}{\sqrt{35}}\sqrt{\frac{65}{99}}\approx\frac{108.54}{5.9161}\times0.8112\approx14.88$

Step2: Calculate z - scores

The z - score formula is $z=\frac{\bar{x}-\mu}{\sigma_{\bar{x}}}$, where $\mu = 129.57$.
For $\bar{x}_1 = 120$, $z_1=\frac{120 - 129.57}{14.88}=\frac{- 9.57}{14.88}\approx - 0.64$.
For $\bar{x}_2 = 140$, $z_2=\frac{140 - 129.57}{14.88}=\frac{10.43}{14.88}\approx0.70$.

Step3: Find the probability

We want $P(120<\bar{X}<140)=P(-0.64 < Z < 0.70)$.
Using the standard normal table, $P(-0.64 < Z < 0.70)=P(Z < 0.70)-P(Z < - 0.64)$.
From the standard - normal table, $P(Z < 0.70)=0.7580$ and $P(Z < - 0.64)=0.2611$.
So $P(-0.64 < Z < 0.70)=0.7580-0.2611 = 0.4969$.

Answer:

$0.4969$