The mean amount spent by a family of four on food per month is $500 with a standard deviation of $75. Assuming that the food costs are normally distributed, what is the probability that a family spends less than $410 per month?
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A. B. C. D. E.Explanation
z = (X - u)/sigma = (410 - 500)/75=-1.2. z = 1.2 is 0.3849. Since 410 is below the mean, 1.0 - 0.8849 = 0.1151.
To solve this problem, we need to use the concept of the standard normal distribution, also known as the Z-distribution. The Z-distribution is a standard normal distribution with a mean of 0 and a standard deviation of 1.
First, we need to standardize the given value of $410 using the formula for the Z-score:
Z = (X - μ) / σ
Where: Z is the Z-score X is the observed value ($410) μ is the mean ($500) σ is the standard deviation ($75)
Plugging in the values, we have:
Z = (410 - 500) / 75 = -1.2
Now, we need to find the probability associated with this Z-score. We can use a Z-table or a statistical calculator to find this probability.
Looking up the Z-table, we find that the area to the left of Z = -1.2 is approximately 0.1151. This means that the probability that a family spends less than $410 per month is 0.1151.
Therefore, the correct answer is A. 0.1151.