Question

Suppose that the distribution for total amounts spent by students vacationing for a week in Florida is normally distributed with a mean of 650 and a standard deviation of 120 . Suppose you take a simple random sample (SRS) of 15 students from this distribution. What is the probability that a SRS of 15 students will spend an average of between 600 and 700 dollars? Round to five decimal places.

Answer 1

Answer:

$P(600<\bar X<700)=0.89347$

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the total amounts spent by students vacationing for a week in Florida of a population, and for this case we know the distribution for X is given by:

$X \sim N(650,120)$  

Where $\mu=650$ and $\sigma=120$

And let $\bar X$ represent the sample mean, the distribution for the sample mean is given by:

$\bar X \sim N(\mu,\frac{\sigma}{\sqrt{n}})$

On this case  $\bar X \sim N(650,\frac{120}{\sqrt{15}})$

We are interested on this probability

$P(600<\bar X<700)$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}$

If we apply this formula to our probability we got this:

$P(600<\bar X<700)=P(\frac{600-\mu}{\frac{\sigma}{\sqrt{n}}}<\frac{X-\mu}{\frac{\sigma}{\sqrt{n}}}<\frac{700-\mu}{\frac{\sigma}{\sqrt{n}}})$

$=P(\frac{600-650}{\frac{120}{\sqrt{15}}}<Z<\frac{700-650}{\frac{120}{\sqrt{15}}})=P(-1.614<Z<1.614)$

And we can find this probability on this way:

$P(-1.614<Z<1.614)=P(Z<1.614)-P(Z<-1.614)$

And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.  

$P(-1.614<Z<1.614)=P(Z<-1.614)-P(Z<-1.614)=0.94674-0.05326=0.89347$

Answer 2

Using the normal distribution and the central limit theorem, it is found that there is a 0.8926 = 89.26% probability that a SRS of 15 students will spend an average of between 600 and 700 dollars.

In a normal distribution with mean $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

• It measures how many standard deviations the measure is from the mean.  
• After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.
• By the Central Limit Theorem, the standard deviation for SRS of n is given by $s = \frac{\sigma}{\sqrt{n}}$

In this problem:

• Mean of 650, thus $\mu = 650$.
• Standard deviation of 120, thus $\sigma = 120$.
• SRS of 15, thus $n = 15, s = \frac{120}{\sqrt{15}}$

The probability is the p-value of Z when X = 700 subtracted by the p-value of Z when X = 600, thus:

X = 700:

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{700 - 650}{\frac{120}{\sqrt{15}}}$

$Z = 1.61$

$Z = 1.61$ has a p-value of 0.9463.

X = 600:

$Z = \frac{X - \mu}{s}$

$Z = \frac{600 - 650}{\frac{120}{\sqrt{15}}}$

$Z = -1.61$

$Z = -1.61$ has a p-value of 0.0537.

0.9463 - 0.0537 = 0.8926

0.8926 = 89.26% probability that a SRS of 15 students will spend an average of between 600 and 700 dollars.

A similar problem is given at https://brainly.com/question/24663213