SOCR EduMaterials Activities Central Limit Theorem Chi square examples
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<math>P(X \ge 1000)= P(Z> \frac{999.5-1050}{17.748}=P(Z>-2.845)=.9977</math> | <math>P(X \ge 1000)= P(Z> \frac{999.5-1050}{17.748}=P(Z>-2.845)=.9977</math> | ||
Below you can see a snapshot for this approximation: | Below you can see a snapshot for this approximation: | ||
- | + | <center>[[Image: SOCR_Activities_CLT_Christou_example2.jpg|600px]]</center> | |
*’’’Answer:’’’ | *’’’Answer:’’’ |
Revision as of 20:48, 14 May 2007
- Answer:
- a. false, the standard deviation of the sample mean is . Thus as the sample size increases, n increases, and as n increases, the standard deviation decreases.
- b. True
- c. False, standard deviation of the sample mean is
- d. True, the standard deviation of the total of a sample of n observations is ; but the standard deviation of the sample mean is Unless n is one, the standard deviation of the total of a sample of n observations exceeds the standard deviation of the sample mean.
- e. False, let's assume σ = 2 and n = 2. In this case, the z-score for would be -2.828 while the z-score for P(X > 4) would be -2. P(Z > − 2.828) > P(Z > − 2). Therefore the statement is false.
- Answer:
- a. Failed to parse (syntax error): P(X \ge 1000)= P(X=1000)+P(X=1001)+....+P(X=1500)= (1500 \choose 1000) \times (.7)^1000 \times (.3)^500 + (1500 \choose 1001) \times (.7)^1001 \times (.3)^499 + ...+(1500 \choose 1500) \times (.7)^1500 \times (.3)^0 = \summation (\1500 \choose X) \times (.7)^X \times (.3)^1500-X
- b. We can use the normal approximation to binomial: and
Below you can see a snapshot for this approximation:
- ’’’Answer:’’’
- a.
Below you can see a snapshot for this part:
- b. ??
- c. The central limit theorem states that the sample mean approaches the normal distribution as the sample size gets bigger. Usually, if we can assume that the sample mean approaches the normal distribution. In this case n = 400. Therefore n satisfies the requirement of a large n.
- d.
- e. In this case, We know that P(T>b) =.975. So now we need to find the 97.5th percentile of this distribution using SOCR. According to the SOCR snapshot below, the 97.5th percentile of this distribution is 3984. Therefore b=3984.
- ’’’Answer:’’’
- a. Failed to parse (unknown function\sqrtn): \overline{X} \sim N(\mu, \frac{\sigma}{\sqrtn}</math. In this case, <math>\overline{X} \sim N(80000, 4518.48)
.
- b. We can find the answer using SOCR. The answer is 0.004032. Please see snapshot below:
- c. We can find the answer right away using SOCR. Please see snapshots below:
55.6% for one hour vs. 86.6% for sample mean. Therefore the sample mean is more likely to be greater than 75000 hours.
- ’’’Answer:’’’
- a. According to the SOCR snapshot below, the 75th percentile is 0.006115.
- b.