Question

the average weekly food expenditure of families in a certain area has a normal distribution with mean rs. 125 and standard deviation rs. 25. what is the probability that a family selected at random from this area will have an average weekly expenditure on food in excess of rs. 175?

Answer 1

To calculate the probability of a family spending more than Rs. 175 on food weekly, a z-score is computed and then used to find the corresponding probability from the standard normal distribution. In this case, the probability is approximately 2.28%.

Step-by-step explanation:

To find the probability that a randomly selected family from the given area has a weekly food expenditure in excess of Rs. 175, you would use the normal distribution. The mean weekly expenditure is Rs. 125, and the standard deviation is Rs. 25.

First, calculate the z-score for Rs. 175 using the formula:

Z = (X - μ) / σ

Where X is Rs. 175, μ (mu) is the mean of Rs. 125, and σ (sigma) is the standard deviation of Rs. 25.

Z = (175 - 125) / 25

Z = 50 / 25

Z = 2

Next, look up this z-score in the standard normal distribution table (or use a calculator), which gives you the probability corresponding to everything below Rs. 175. To find the probability of expenditure exceeding this amount, subtract this value from 1.

If, for example, the table gives a probability of 0.9772 for a z-score of 2, then the probability of a family spending more than Rs. 175 is:

1 - 0.9772 = 0.0228

So, there is approximately a 2.28% chance that a randomly selected family will spend more than Rs. 175 weekly on food.