4. Meteorologist predicted that sea level in a specific place will raise by 20 c

Question

4. Meteorologist predicted that sea level in a specific place will raise by 20 cm over the normal level. We assume that that sea level over the normal level in this flood, X, is normally distributed: X ~ N(20, ) . Assume that standard deviation 4,5 cm Flood was 10 cm lower than expected. What is is the probability of flood being 10 cm or more under the expected level? Find 95% spreading interval for water level. Draw a simple figure. Meteorologist wants to make a prognosis for water level: "k cm". Probability that water level in the flood is smaller than k should be at maximum equal to 0,05. What is the prognosis k? Assume that expected water level is unknown and equal to . What do we need to set up a 95% confidence interval for the expected water level? a. b. c. d.

Explanation / Answer

Answer to all parts down below:

a. P(X<=10) = P(Z<= 10-20/4.5) = P(Z<= -2.22) = .0132

b. The 95% confidence level is given by : Mean +/- Z*Sigma = 20 +/- 1.96*4.5 = 11.18 to 28.82

c. P(X<=k) = .05, So, Z for this 0.05 is -1.645

So, k = 20 + 1.96*-1.645 = 16.7758

d. When the pop. mean is unknown this part is asking us to assume Mean as Mu.  So, to fund out the 95% CI we have : Mu +/- 1.96*Sigma = 20 +/- 1.96*-1.645 = 16.7758 to 23.2242

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