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

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

We are given;

Sample size; n = 25

Population mean; μ = 650

Population standard deviation; σ = 120

We see that the sample size is less than 30, but from the Central Limit Theorem, we can say that the sampling distribution of the sample means with size n can be approximated to a normal distribution. Thus, test statistic formula is;

z = (x⁻ - μ)/(σ/√n)

Now, the probability that a SRS of 25 students will spend an average of between 600 and 700 dollars will be the difference between the p-values when; x⁻ = 600 and when x⁻ = 700

when x⁻ = 600;

z = (600 - 650)/(120/√25)

z = 2.08

when x⁻ = 700;

z = (700 - 650)/(120/√25)

z = -2.08

From z-score tables;

p-value at z = 2.08 is p = 0.9812

p-value at z = -2.08 is p = 0.0188

Thus, probability that a SRS of 25 students will spend an average of between 600 and 700 dollars is; 0.9812 - 0.0188 = 0.9624 or 96.24%

Thus, in conclusion, the probability that a SRS of 25 students will spend an average of between 600 and 700 dollars

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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 and standard deviation , the zscore of a measure X is given by:

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 and standard deviation , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean and standard deviation .

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:

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

By the Central Limit Theorem

has a pvalue of 0.9812

X = 600

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

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step-by-step explanation: