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 25 students from this distribution. What is the probability that a SRS of 25 students will spend an average of between 600 and 700 dollars

Answer 1

Given Information:  

Mean = μ = 650  

Standard deviation = σ = 120

Number of students = n = 25

Required Information:  

P( 600 < X < 700) = ?  

Answer:  

P( 600 < X < 700) = 0.9625 = 96.25%

Explanation:  

Let random variable X denotes total amount spent by students vacationing for a week in Florida.

The probability that a SRS of 25 students will spend an average of between 600 and 700 dollars is given by

P( 600 < X < 700) = P((x - μ)/σ/√n < z < (x - μ)/σ/√n)  

P( 600 < X < 700) = P((600 - 650)/120/√25 < z < (700 - 650)/120/√25)  

P( 600 < X < 700) = P( -2.08 < z < 2.08)

P( 600 < X < 700) = P( z < 2.08) - P( z > -2.08)

The z-score from the z-table yields,

P( 600 < X < 700) = 0.98124 - 0.01876

P( 600 < X < 700) = 0.9625

P( 600 < X < 700) = 96.25%

Therefore, there is 96.25% probability that a SRS of 25 students will spend an average of between 600 and 700 dollars.

Answer 2

Answer:

96.24% probability that a SRS of 25 students will spend an average of between 600 and 700 dollars

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

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

The Z-score 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. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

$\mu = 650, \sigma = 120, n = 25, s = \frac{120}{\sqrt{25}} = 24$

What is the probability that a SRS of 25 students will spend an average of between 600 and 700 dollars

This is the pvalue of Z when X = 700 subtracted by the pvalue of Z when X = 600. So

X = 700

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

By the Central Limit Theorem

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

$Z = \frac{700 - 650}{24}$

$Z = 2.08$

$Z = 2.08$ has a pvalue of 0.9812

X = 600

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

$Z = \frac{600 - 650}{24}$

$Z = -2.08$

$Z = -2.08$ has a pvalue of 0.0188

0.9812 - 0.0188 = 0.9624

96.24% probability that a SRS of 25 students will spend an average of between 600 and 700 dollars