Question

the monthly electric bills in a certain community are normally distributed, with a mean of $245 and a standard distribution of $35. if there are 481 households in the community, about how many of them have electric bills less than $200? table shows values to the left of the z - score

z	0.00	0.01	0.02	0.03	0.04	0.05	0.06	0.07
0.7	0.7580	0.7611	0.7642	0.7673	0.7704	0.7734	0.7764	0.7794
0.8	0.7881	0.7910	0.7939	0.7967	0.7995	0.8023	0.8051	0.8078
0.9	0.8159	0.8186	0.8212	0.8238	0.8264	0.8289	0.8315	0.8340
1.0	0.8413	0.8438	0.8461	0.8485	0.8508	0.8531	0.8554	0.8577
1.1	0.8643	0.8665	0.8686	0.8708	0.8729	0.8749	0.8770	0.8790
-1.1	0.1357	0.1335	0.1314	0.1292	0.1271	0.1251	0.1230	0.1210
-1.0	0.1587	0.1562	0.1539	0.1515	0.1492	0.1469	0.1446	0.1423
-0.9	0.1841	0.1814	0.1788	0.1762	0.1736	0.1711	0.1685	0.1660
-0.8	0.2119	0.2090	0.2061	0.2033	0.2005	0.1977	0.1949	0.1922
-0.7	0.2420	0.2389	0.2358	0.2327	0.2296	0.2266	0.2236	0.2206
-0.6	0.2743	0.2709	0.2676	0.2643	0.2611	0.2578	0.2546	0.2514

a. 21
b. 45
c. 81
d. 85

Explanation:

Step1: Calculate the z - score

The formula for the z - score is $z=\frac{x-\mu}{\sigma}$, where $x = 200$, $\mu=245$, and $\sigma = 35$.
$z=\frac{200 - 245}{35}=\frac{- 45}{35}\approx - 1.29$

Step2: Find the proportion from the z - table

Looking up $z=-1.29$ in the z - table (values to the left of the z - score), we find the proportion of values to the left of $z=-1.29$. Since the table doesn't have exactly $-1.29$, we take the average of the values for $z=-1.28$ and $z=-1.29$. The value for $z = - 1.28$ is $0.1003$ and for $z=-1.29$ is $0.0985$. The average is $\frac{0.1003 + 0.0985}{2}=0.0994$.

Step3: Calculate the number of households

Multiply the proportion by the total number of households. There are $n = 481$ households.
The number of households with electric bills less than $200$ is $0.0994\times481\approx481\times0.1 = 48.1\approx45$ (closest value in the options)

Answer:

B. 45