SOCR EduMaterials Activities Central Limit Theorem Chi square examples

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Answer:
a. false, the standard deviation of the sample mean is $\frac{\sigma}{\sqrt n}$ . 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 $\frac{\sigma}{\sqrt n}$
d. True, the standard deviation of the total of a sample of n observations is $n\sqrt \sigma$ ; but the standard deviation of the sample mean is $\frac{\sigma}{\sqrt n}.$ 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 $P(\overline{X} > 4)$ 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: $\mu = np = 1500 \times 0.70 = 1050.$ and $\sigma = \sqrt npq = \sqrt1500 \times 0.7 \times 0.3= 17.748.$

$P(X \ge 1000)= P(Z> \frac{999.5-1050}{17.748}=P(Z>-2.845)=.9977$

Answer:
a. $\overline{X} \sim N(8, \frac{20}{\sqrt400}). P(\overline{X} <6.50) =P(Z<-1.5)=.0667$
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 $n \ge 30$ 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. $\overline{X} \sim N(8,1).$ According to the snapshot below, the middle 80% of this distribution is (6.721,9.279). Therefore $w = 8 - 6.721 = 1.29$

e. $T \sim N(n\mu,\sigma\sqrt n).$ In this case, $T \sim N(3200,400).$ 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. $\overline{X} \sim N(\mu, \frac{\sigma}{\sqrt n }$ . In this case, $\overline{X} \sim N(80000, 4518.48)$ .
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.