Respuesta :
Answer:
(a) The distribution of X is N (9.8, 0.8²).
(b) The probability that an American consumes between 8.8 and 9.9 grams of sodium per day is 0.4461.
(c) The middle 30% of American men consume between 9.5 grams to 10.1 grams of sodium.
Step-by-step explanation:
The random variable X is defined as the amount of sodium consumed.
The random variable X has an average value of, μ = 9.8 grams.
The standard deviation of X is, σ = 0.8 grams.
(a)
It is provided that the sodium consumption of American men is normally distributed.
The random variable X follows a normal distribution with parameters μ = 9.8 grams and σ = 0.8 grams.
Thus, the distribution of X is N (9.8, 0.8²).
(b)
If X ~ N (µ, σ²), then [tex]Z=\frac{X-\mu}{\sigma}[/tex], is a standard normal variate with mean, E (Z) = 0 and Var (Z) = 1. That is, Z ~ N (0, 1).
To compute the probability of Normal distribution it is better to first convert the raw score (X) to z-scores.
Compute the probability that an American consumes between 8.8 and 9.9 grams of sodium per day as follows:
[tex]P(8.8<X<9.9) =P(\frac{8.8-9.8}{0.8}<\frac{X-\mu}{\sigma}<\frac{9.9-9.8}{0.8})[/tex]
[tex]=P(-1.25<Z<0.125)\\=P(Z<0.125)-P(Z<-1.25)\\=0.5517-0.1056\\=0.4461[/tex]
Thus, the probability that an American consumes between 8.8 and 9.9 grams of sodium per day is 0.4461.
(c)
The probability representing the middle 30% of American men consuming sodium between two weights is:
[tex]P(x_{1}<X<x_{2})=0.30\\\Rightarrow P(-z<Z<z)=0.30[/tex]
Compute the value of z as follows:
[tex]P(-z<Z<z)=0.30\\P(Z<z)-P(Z<-z)=0.30\\P(Z<z)-[1-P(Z<z)]=0.30\\2P(Z<z)-1=0.30\\P(Z<z)=0.65[/tex]
The value of z for P (Z < z) = 0.65 is 0.39.
Compute the value of x₁ and x₂ as follows:
[tex]-z=\frac{x_{1}-\mu}{\sigma}\\-0.39=\frac{x_{1}-9.8}{0.8}\\x_{1}=9.8-(0.39\times 0.8)\\x_{1}=9.488\\x_{1}\approx9.5[/tex] [tex]z=\frac{x_{2}-\mu}{\sigma}\\0.39=\frac{x_{1}-9.8}{0.8}\\x_{1}=9.8+(0.39\times 0.8)\\x_{1}=10.112\\x_{1}\approx10.1[/tex]
Thus, the middle 30% of American men consume between 9.5 grams to 10.1 grams of sodium.
For the given context, the figures for consumption of sodium by American men per day are:
- X is normally distributed with mean 9.8 grams and standard deviation 0.8 grams, or [tex]X \sim N(9.8, 0.8)[/tex]
- [tex]P(8.8 < X< 9.9) \approx[/tex]
- The middle 30% of American men consume sodium between 8.768 grams to 10.832 grams approx.
How to get the z scores?
If we've got a normal distribution, then we can convert it to standard normal distribution and its values will give us the z score.
If we have
[tex]X \sim N(\mu, \sigma)[/tex]
(X is following normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex])
then it can be converted to standard normal distribution as
[tex]Z = \dfrac{X - \mu}{\sigma}, \\\\Z \sim N(0,1)[/tex]
(Know the fact that in continuous distribution, probability of a single point is 0, so we can write
[tex]P(Z \leq z) = P(Z < z) )[/tex]
Also, know that if we look for Z = z in z tables, the p value we get is
[tex]P(Z \leq z) = \rm p \: value[/tex]
For the given case, we have:
X = random variable tracking the amount of sodium consumed by american men per day
Thus, as average consumption is of 9.8 grams with standard deviation 0.8 grams, we get: [tex]X \sim N(\mu = 9.8, \sigma = 0.8)[/tex] (since it is given to be normally distributed)
The probability that an American man consumes between 8.8 and 9.9 grams of sodium per day is [tex]P(8.8 < X < 9.9)[/tex]
This can be rewritten as:
[tex]P(8.8 < X < 9.9) = P(X < 9.9) - P( X < 8.8)[/tex]
Using the conversion to standard normal distribution, the needed probability is:
[tex]P(8.8 < X < 9.9) = P(X < 9.9) - P( X < 8.8)\\\\P(8.8 < X < 9.9) = P(Z < \dfrac{9.9-9.8}{0.8}= 0.125)} - P(Z < \dfrac{8.8 - 9.8}{0.8} =-1.25)}\\\\P(8.8 < X < 9.9) = P(Z < 0.125) - P(Z < -1.25)[/tex]
Using the z-tables, we get the p-value for Z =0.125 as 0.5478 and for Z = -1.25 as 0.1056
Thus, the needed probability is :
[tex]P(8.8 < X < 9.9) = P(Z < 0.125) - P(Z < -1.25) \approx 0.5478 - 0.1056 \\P(8.8 < X < 9.9) \approx 0.4422[/tex]
Also, since the normal distribution is symmetric , thus, the mid 30% is 15% on left of mean of X's distribution, and 15% on right, with outer limits being equidistant from the mean, let we say it is 'd' units from left and right of mean, then:
[tex]P(\mu - d < X < \mu + d) = 30\% = 0.3\\P(X < \mu + d) - P(X < \mu - d) = 0.3\\P(Z < \dfrac{\mu + d - \mu}{\sigma}) - P(Z < \dfrac{\mu - d - \mu}{\sigma}) = 0.3\\P(Z < \dfrac{d}{0.8}) - P(Z < - \dfrac{d}{0.8}) = 0.3\\\\P(Z < \dfrac{d}{0.8}) - (1 - P(Z < \dfrac{d}{0.8})) = 0.8\\\\2P(Z < \dfrac{d}{0.8}) - 1 = 0.8\\\\P(Z < \dfrac{d}{0.8}) = 0.9[/tex]
(it is because, if Z is standard normal variate then P(Z < -a) = 1 - P(Z < a) )
From z tables, we see that for Z = 1.29 , the p value comes out as 0.9, thus,
[tex]\dfrac{d}{0.8} \approx 1.29\\\\d \approx 1.032[/tex]
Thus, the mid-interval in which 30% of American men consuming sodium comes as: [9.8 - d, 9.8 + d] = [8.768, 10.832](in grams) approx.
So, middle 30% of American men consume sodium between 8.768 grams to 10.832 grams approx.
Thus, the figures for consumption of sodium by American men per day are:
- X is normally distributed with mean 9.8 grams and standard deviation 0.8 grams, or [tex]X \sim N(9.8, 0.8)[/tex]
- [tex]P(8.8 < X< 9.9) \approx[/tex]
- The middle 30% of American men consume sodium between 8.768 grams to 10.832 grams approx.
Learn more about standard normal distribution here:
https://brainly.com/question/10984889
