Respuesta :
Answer:
a) The sampling distributio will be normal with mean 600 and standard deviation 11.07.
b) - 68% of the samples will have means between 589 and 611 apples per tree.
- 95% of the samples will have means between 578 and 622 apples per tree.
- 99.7% of the samples will have means between 567 and 633 apples per tree.
c) The probability that a random sample of 40 trees has an average amount of apples of 620 pounds or higher is 0.035.
Step-by-step explanation:
a) The sampling distribution for a population with mean of 600 pounds and standard deviation of 70 pounds, with a sample size of 40 trees, will have the following parameters:
[tex]\mu_s=\mu=600\\\\ \sigma_s=\dfrac{\sigma}{\sqrt{n}}=\dfrac{70}{\sqrt{40}}=\dfrac{70}{6.325}=11.07[/tex]
b) Using the 68-95-99.7 rule for the sampling distribution, we can tell that:
- 68% of the samples will have means between 589 and 611 apples per tree.
[tex]X_1=\mu+z_1\cdot\sigma/\sqrt{n}=600-1\cdot 70\sqrt{40}=600-11=589\\\\X_2=\mu+z_2\cdot\sigma/\sqrt{n}=600+1\cdot 70\sqrt{40}=600+11=611[/tex]
- 95% of the samples will have means between 578 and 622 apples per tree.
[tex]X_1=\mu+z_1\cdot\sigma/\sqrt{n}=600-2\cdot 70\sqrt{40}=600-22=578\\\\X_2=\mu+z_2\cdot\sigma/\sqrt{n}=600+2\cdot 70\sqrt{40}=600+22=622[/tex]
- 99.7% of the samples will have means between 567 and 633 apples per tree.
[tex]X_1=\mu+z_1\cdot\sigma/\sqrt{n}=600-3\cdot 70\sqrt{40}=600-33=567\\\\X_2=\mu+z_2\cdot\sigma/\sqrt{n}=600+3\cdot 70\sqrt{40}=600+33=633[/tex]
c) We can calculate the probability with the z-score for X=620.
[tex]z=\dfrac{X-\mu}{\sigma/\sqrt{n}}=\dfrac{620-600}{70/\sqrt{40}}=\dfrac{20}{11.068}=1.807\\\\\\P(X>620)=P(z>1.807)=0.035[/tex]
The probability that a random sample of 40 trees has an average amount of apples of 620 pounds or higher is 0.035.
