Expand the given power using the Binomial Theorem. (z – 11)4
2 answers:
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Answer:
One can be 99% confident the true mean shell length lies within the above interval.
The population has a relative frequency distribution that is approximately normal.
Step-by-step explanation:
We are given that Time-depth recorders were deployed on 6 of the 76 captured turtles. These 6 turtles had a mean shell length of 51.3 cm and a standard deviation of 6.6 cm.
The pivotal quantity for a 99% confidence interval for the true mean shell length is given by;
P.Q. = ~
where, = sample mean shell length = 51.3 cm
s = sample standard deviation = 6.6 cm
n = sample of turtles = 6
= true mean shell length
Now, the 99% confidence interval for =
Here, = 1% so
= 0.5%. So, the critical value of t at 0.5% significance level and 5 (6-1) degree of freedom is 4.032.
<u>So, 99% confidence interval for</u> =
= [51.3 - 10.864 , 51.3 + 10.864]
= [40.44 cm, 62.16 cm]
The interpretation of the above result is that we are 99% confident that the true mean shell length lie within the above interval of [40.44 cm, 62.16 cm].
The assumption about the distribution of shell lengths must be true in order for the confidence interval, part a, to be valid is that;
C. The population has a relative frequency distribution that is approximately normal.
This assumption is reasonably satisfied as the data comes from the whole 76 turtles and also we don't know about population standard deviation.
Answer: $2193
Step-by-step explanation:
v(x) = 32,500
Plug two and three in for x
v(2) = 32,500
v(2) = 32,500(.8464)
v(2) = 27,500
v(3) = 32,500
v(3) = 32,500(.778688)
v(3) = 25,307.36
Subtract v(3) from v(2)
27,500 - 25,307.36 = 2192.64
Round to the nearest dollar to get $2193