Question

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? 0.2158 0.8750 0.0362 0.1151

Answer 1

                                  
Are you familiar with z-scores?  According to the definition,

       (given numerical value) - (mean)
z = ---------------------------------------------
              standard deviation

Thus, with the given numerical value equal to 410 and the std. dev. equal to 75, the corresponding z-score is 
       
          410-500                  -90
z = --------------------- = --------------- = -1.2
               75                        75

Use a table of z-scores to determine the area under the standard normal curve to the left of z = -1.2.  Your result is the probability that a given family chosen at random spends less than $410 per month.

Answer 2

Answer:

0.1151

Step-by-step explanation:

Problems of normally distributed samples can be 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.

In this problem, we have that:

$\mu = 500, \sigma = 75$.

What is the probability that a family spends less than $410 per month?

This probability is the pvalue of Z when X = 410. So:

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

$Z = \frac{410 - 500}{75}$

$Z = -1.2$

$Z = -1.2$ has a pvalue of 0.1151. This is the answer.