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. Let X X

Answer 1

Answer:

a) $\mu -3\sigma = 336-3*6=318$

$\mu+-3\sigma = 336+3*6=354$

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 $X\sim N(\mu = 336. \sigma =6)$ 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:

$\mu -3\sigma = 336-3*6=318$

$\mu+-3\sigma = 336+3*6=354$

(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%