What number is equivalent to 4.054

In the third week of July, a random sample of 40 farming regions gave a sample mean of x = $6.88 per 100 pounds of watermelon. A

suter [353]

Answer:

a. [6.6350,7.3950]

b. ME=0.5150

Step-by-step explanation:

a. Given that n=40, $\bar x=6.88$ and that: $z_{\alpha/2}=z_{0.05}=1.645$

The required 90% confidence interval can be calculated as:

$\bar x\pm(margin \ of \ error)\\\\\bar x\pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}\\\\6.88\pm(1.645\times \frac{1.98}{\sqrt{40}})\\\\=[6.3650,7.3950]$

Hence, the 90% confidence interval for the population mean cash value of this crop is [6.6350,7.3950]

b. The margin of error at 90% confidence interval is calculated as:

$ME= z_{\alpha/2}\frac{\sigma}{\sqrt{n}}\\\\=(1.645\times \frac{1.98}{\sqrt{40}})\\\\=0.5150$

Hence, the margin of error is 0.5150

8 0

4 years ago

Stacy uses a spinner with six equal sections numbered 2, 2, 3, 4, 5, and 6 to play a game. Write a probability model for this ex

allsm [11]

Answer:

We estimate to have 8.33 times the number 6 in 50 trials.

Step-by-step explanation:

Let us consider a success to get a 6. In this case, note that the probability of having a 6 in one spin is 1/6. We can consider the number of 6's in 50 spins to be a binomial random variable. Then, let X to be the number of trials we get a 6 out of 50 trials. Then, we have the following model.

$P(X=k) = \binom{50}{k}(\frac{1}{6})^k(\frac{5}{6})^{50-k}$