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 $95. what is the approximate percentage of buyers who paid between $1105 and $1200? what is the approximate percentage of buyers who paid between $1200 and $1390? what is the approximate percentage of buyers who paid between $1200 and $1485? what is the approximate percentage of buyers who paid more than $1390? what is the approximate percentage of buyers who paid more than $1485? what is the approximate percentage of buyers who paid between $1105 and $1295?

Explanation:

Step1: Recall the empirical rule for normal distribution

The empirical rule states that:

• Approximately \(68\%\) of the data lies within \(1\) standard deviation (\(\mu\pm\sigma\)) of the mean.
• Approximately \(95\%\) of the data lies within \(2\) standard deviations (\(\mu\pm2\sigma\)) of the mean.
• Approximately \(99.7\%\) of the data lies within \(3\) standard deviations (\(\mu\pm3\sigma\)) of the mean.

Since the normal distribution is symmetric about the mean \(\mu = 1200\) and \(\sigma=95\).

Step2: Calculate the number of standard deviations for each value

For \(x = 1105\), \(z=\frac{1200 - 1105}{95}=1\) (left of the mean).
For \(x = 1295\), \(z=\frac{1295 - 1200}{95}=1\) (right of the mean).
For \(x = 1390\), \(z=\frac{1390 - 1200}{95}=2\) (right of the mean).
For \(x = 1485\), \(z=\frac{1485 - 1200}{95}=3\) (right of the mean).

Step3: Calculate the percentage for each interval

• Between \(1105\) and \(1200\):

Since the data within \(\mu\pm\sigma\) is \(68\%\), and the distribution is symmetric. The percentage between \(\mu-\sigma\) and \(\mu\) is \(\frac{68\%}{2}=34\%\).

• Between \(1200\) and \(1390\):

The data within \(\mu\pm2\sigma\) is \(95\%\). The data within \(\mu\) and \(\mu + 2\sigma\) is \(\frac{95\%}{2}=47.5\%\).

• Between \(1200\) and \(1485\):

The data within \(\mu\pm3\sigma\) is \(99.7\%\). The data within \(\mu\) and \(\mu+3\sigma\) is \(\frac{99.7\%}{2}=49.85\%\approx50\%\).

• More than \(1390\):

The data within \(\mu\pm2\sigma\) is \(95\%\). The data more than \(\mu + 2\sigma\) is \(\frac{100\% - 95\%}{2}=2.5\%\).

• More than \(1485\):

The data within \(\mu\pm3\sigma\) is \(99.7\%\). The data more than \(\mu+3\sigma\) is \(\frac{100\% - 99.7\%}{2}=0.15\%\approx0.1\%\).

• Between \(1105\) and \(1295\):

The data within \(\mu\pm\sigma\) is \(68\%\).

Answer:

• Between \(1105\) and \(1200\): \(34\%\)
• Between \(1200\) and \(1390\): \(47.5\%\)
• Between \(1200\) and \(1485\): \(50\%\)
• More than \(1390\): \(2.5\%\)
• More than \(1485\): \(0.1\%\)
• Between \(1105\) and \(1295\): \(68\%\)