If t is any real number, prove that 1+(tant)^2=(sect)^2
1 answer:
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Answer:
a)
b) For this case we know that within 1 deviation from the mean we have 68% of the data, and on the tails we need to have 100-68 =32% of the data with each tail with 16%. The value 342 is above the mean one deviation so then we need to have accumulated below this value 68% +(100-68)/2 = 68%+16% =84%
And then the % above would be 100-84= 16%
Step-by-step explanation:
Assuming this question : "Bigger animals tend to carry their young longer before birth. The length of horse pregnancies from conception to birth varies according to a roughly normal distribution with mean 336 days and standard deviation 6 days. Use the 68-95-99.7 rule to answer the following questions. "
(a) Almost all (99.7%) horse pregnancies fall in what range of lengths?
First we need to remember the concept of empirical rule.
From this case we assume that where X represent the random variable "length of horse pregnancies from conception to birth"
The empirical rule, also referred to as the three-sigma rule or 68-95-99.7 rule, is a statistical rule which states that for a normal distribution, almost all data falls within three standard deviations (denoted by σ) of the mean (denoted by µ). Broken down, the empirical rule shows that 68% falls within the first standard deviation (µ ± σ), 95% within the first two standard deviations (µ ± 2σ), and 99.7% within the first three standard deviations (µ ± 3σ).
From the empirical rule we know that we have 99.7% of the data within 3 deviations from the mean so then we can find the limits for this case with this:
(b) What percent of horse pregnancies are longer than 342 days?
For this case we know that within 1 deviation from the mean we have 68% of the data, and on the tails we need to have 100-68 =32% of the data with each tail with 16%. The value 342 is above the mean one deviation so then we need to have accumulated below this value 68% +(100-68)/2 = 68%+16% =84%
And then the % above would be 100-84= 16%