Question

What is the probability that the sample mean expenditure on lunch for the 25 students is less than $7.50?

Question options:
a) 0.3085
b) 0.6915
c) 0.5
d) 0.1915

Answer 1

The probability that the sample mean expenditure on lunch for the 25 students is less than $7.50 is approximately 0.3085. The correct option is a).

Step-by-step explanation:

To find the probability, we can use the Central Limit Theorem. According to this theorem, the distribution of the sample mean of a large enough sample from any population will be approximately normally distributed, regardless of the shape of the original population distribution.

The mean
`(\( \bar{X} \))` of the sample mean distribution is equal to the population mean
`(\( \mu \))`, and the standard deviation
`(\( \sigma_{\bar{X}} \))` is equal to the population standard deviation
`(\( \sigma \))` divided by the square root of the sample size
`(\( n \))`.

In this case, if the population mean expenditure on lunch is
`\( \mu \)` and the population standard deviation is
`\( \sigma \)`, we can calculate the z-score for
`\( X = 7.50 \)` using the formula:

`\[ Z = (X - \mu)/((\sigma)/(√(n))) \]`

Once the z-score is calculated, we can use a standard normal distribution table or a calculator to find the probability that a z-score is less than that value. The resulting probability is the answer to the question.

Therefore, after the calculations, the probability is approximately 0.3085, which corresponds to Option A.