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| - | *'''Answer:'''
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| - | *a. false, the standard deviation of the sample mean is <math>\frac{\sigma}{\sqrt n}</math>. Thus as the sample size increases, n increases, and as n increases, the standard deviation decreases.
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| - | *b. True
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| - | *c. False, standard deviation of the sample mean is <math>\frac{\sigma}{\sqrt n}</math>
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| - | *d. True, the standard deviation of the total of a sample of n observations is <math>n\sqrt \sigma</math>; but the standard deviation of the sample mean is <math>\frac{\sigma}{\sqrt n}.</math>Unless n is one, the standard deviation of the total of a sample of n observations exceeds the standard deviation of the sample mean.
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| - | *e. False, let's assume <math>\sigma=2</math> and <math>n=2</math>. In this case, the z-score for <math>P(\overline{X} > 4)</math> would be -2.828 while the z-score for <math>P(X>4)</math> would be -2. <math>P(Z>-2.828) > P(Z>-2) </math>. Therefore the statement is false.
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| - | *'''Answer:'''
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| - | *a. <math>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</math>
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| - | *b. We can use the normal approximation to binomial: <math>\mu = np = 1500 \times 0.70 = 1050.</math> and <math>\sigma =
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| - | \sqrt npq = \sqrt1500 \times 0.7 \times 0.3= 17.748.</math>
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| - | <math>P(X \ge 1000)= P(Z> \frac{999.5-1050}{17.748}=P(Z>-2.845)=.9977</math>
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| - | * Below you can see a snapshot for this approximation:
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| - | <center>[[Image: SOCR_Activities_CLT_Christou_example2.jpg|600px]]</center>
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| - | *'''Answer:'''
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| - | *a. <math>\overline{X} \sim N(8, \frac{20}{\sqrt400}). P(\overline{X} <6.50) =P(Z<-1.5)=.0667</math>
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| - | *Below you can see a snapshot for this part:
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| - | <center>[[Image: SOCR_Activities_CLT_Christou_example3_a.jpg|600px]]</center>
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| - | *b. ??
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| - | *c. The central limit theorem states that the sample mean approaches the normal distribution as the sample size gets bigger. Usually, if <math> n \ge 30</math>we can assume that the sample mean approaches the normal distribution. In this case <math>n=400</math>. Therefore n satisfies the requirement of a large n.
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| - | *d.<math>\overline{X} \sim N(8,1).</math>According to the snapshot below, the middle 80% of this distribution is (6.721,9.279). Therefore <math>w=8-6.721 =1.29</math>
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| - | <center>[[Image: SOCR_Activities_Normal_Christou_example3_d.jpg|600px]]</center>
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| - | *e. <math>T \sim N(n\mu,\sigma\sqrt n).</math> In this case, <math>T \sim N(3200,400).</math> We know that <math>P(T>b) =.975.</math>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.
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| - | <center>[[Image: SOCR_Activities_Normal_Christou_example3_e.jpg|600px]]</center>
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| - | *’’’Answer:’’’
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| - | *a. <math>\overline{X} \sim N(\mu, \frac{\sigma}{\sqrt n }</math>. In this case, <math>\overline{X} \sim N(80000, 4518.48)</math>.
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| - | *b. We can find the answer using SOCR. The answer is 0.004032. Please see snapshot below:
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| - | <center>[[Image: SOCR_Activities_Normal_Christou_example4_.jpg|600px]]</center>
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| - | *c. We can find the answer right away using SOCR. Please see snapshots below:
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| - | <center>This is the distribution for <math>X</math></center>
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| - | <center>[[Image: SOCR_Activities_Normal_Christou_example4_c.jpg|600px]]</center>
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| - | <center>This is the distribution for <math>\overline{X}</math></center>
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| - | <center>[[Image: SOCR_Activities_Normal_Christou_example3_c2.jpg|600px]]</center>
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| - | *The probabilities are 55.6% for one hour vs. 86.6% for sample mean. Therefore the sample mean is more likely to be greater than 75000 hours.
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| - | *’’’Answer:’’’
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| - | *a. According to the SOCR snapshot below, the 75th percentile is 0.006115.
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| - | <center>[[Image: SOCR_Activities_Normal_Christou_example5_aa.jpg|600px]]</center>
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| - | *b.
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