Question

In: Statistics and Probability

Consider eruptions of Old Faithful geyser in Yellowstone National Park. The distribution of eruptions is said...

Consider eruptions of Old Faithful geyser in Yellowstone National Park. The distribution of eruptions is said to be roughly normal and based on a study done in the park in 2011 has a mean time between eruptions of 93 minutes with a standard deviation of 9.5 minutes.

A)Suppose we are still studying 60 eruptions and we want to know what the longest 5% of mean times between eruptions would be. What average time would put us in the longest 5%?

B)How would your previous answer change if we were only studying 20 eruptions?

C)If we study 45 eruptions, what is the chance that a mean eruption time would differ from the true mean by less than 5 minutes?

D)If we study 45 eruptions, what is the chance that a mean eruption time would differ from the true mean by less than 1 minute?

Solutions

Expert Solution

a)

mean μ= 93
standard deviation σ= 9.5000
sample size       =n= 60
std error=σ=σ/√n= 1.226

as longest 5% will fall on 95th percentile ; therefore

for 95th percentile critical value of z= 1.645
therefore corresponding value=mean+z*std deviation= 95.02 minutes

b)

sample size       =n= 20
std error=σ=σ/√n= 2.124
for 95th percentile critical value of z= 1.645
therefore corresponding value=mean+z*std deviation= 96.49 minute

c)

for normal distribution z score =(X-μ)/σx
here mean=       μ= 93
std deviation   =σ= 9.5000
sample size       =n= 45
std error=σ=σ/√n= 1.4162
probability = P(88<X<98) = P(-3.53<Z<3.53)= 0.9998-0.0002= 0.9996

d)

probability = P(92<X<94) = P(-0.71<Z<0.71)= 0.7611-0.2389= 0.5222

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