Question

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?

Answer 1

a)

mean μ=	93
standard deviation σ=	9.5000
sample size       =n=	60
std error=σx̅=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=σx̅=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=σx̅=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