Question

the graph illustrates a normal distribution for the prices paid for a particular model of hd television. the mean price paid is $1200 and the standard deviation is $60.
1020 1080 1140 1200 1260 1320 1380
distribution of prices
what is the approximate percentage of buyers who paid less than $1020?
%
what is the approximate percentage of buyers who paid between $1200 and $1320?
%
what is the approximate percentage of buyers who paid between $1140 and $1200?
%
what is the approximate percentage of buyers who paid between $1200 and $1380?
%
what is the approximate percentage of buyers who paid less than $1080?
%
what is the approximate percentage of buyers who paid between $1140 and $1260?
%

Explanation:

Step1: Recall the properties of the normal distribution

In a normal distribution, about 68% of the data lies within 1 - standard - deviation of the mean ($\mu\pm\sigma$), about 95% lies within 2 - standard - deviations ($\mu\pm2\sigma$), and about 99.7% lies within 3 - standard - deviations ($\mu\pm3\sigma$). Here, $\mu = 1200$ and $\sigma=60$.

Step2: Calculate the z - scores and find the percentages

1. For the price $x = 1020$, the z - score is $z=\frac{x - \mu}{\sigma}=\frac{1020 - 1200}{60}=\frac{- 180}{60}=-3$. The percentage of data less than $z=-3$ is approximately 0.15%.
2. For the range $1200$ to $1320$, $\mu = 1200$, $x = 1320$, $z=\frac{1320 - 1200}{60}=2$. The percentage of data between $\mu$ and $\mu + 2\sigma$ is $\frac{95\%}{2}=47.5\%$.
3. For the range $1140$ to $1200$, $\mu = 1200$, $x = 1140$, $z=\frac{1140 - 1200}{60}=-1$. The percentage of data between $\mu-\sigma$ and $\mu$ is $\frac{68\%}{2}=34\%$.
4. For the range $1200$ to $1380$, $\mu = 1200$, $x = 1380$, $z=\frac{1380 - 1200}{60}=3$. The percentage of data between $\mu$ and $\mu + 3\sigma$ is $\frac{99.7\%}{2}=49.85\%$.
5. For the price $x = 1080$, $z=\frac{1080 - 1200}{60}=-2$. The percentage of data less than $z = - 2$ is $\frac{100\% - 95\%}{2}=2.5\%$.
6. For the range $1140$ to $1260$, $\mu = 1200$, $x_1 = 1140$, $z_1=\frac{1140 - 1200}{60}=-1$, $x_2 = 1260$, $z_2=\frac{1260 - 1200}{60}=1$. The percentage of data between $\mu-\sigma$ and $\mu+\sigma$ is 68%.

Answer:

1. 0.15
2. 47.5
3. 34
4. 49.85
5. 2.5
6. 68