Question

the graph illustrates a normal distribution for the prices paid for a particular model of hd television. the mean price paid is $1800 and the standard deviation is $95. round answers to 2 decimal places, use technology. what is the probability that a buyer paid between $1705 and $1895? what is the probability that a buyer paid between $1610 and $1800? what price would the buyer pay to get 9% the most expensive hd televisions?

Explanation:

Step1: Calculate z - scores

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(\mu = 1800\) (mean) and \(\sigma=95\) (standard deviation).
For \(x = 1705\), \(z_1=\frac{1705 - 1800}{95}=\frac{-95}{95}=- 1\)
For \(x = 1895\), \(z_2=\frac{1895 - 1800}{95}=\frac{95}{95}=1\)
Using a standard normal table or technology (e.g., the normalcdf function on a TI - 84: normalcdf\((-1,1)\)), the probability \(P(-1<Z<1)\)
For \(x = 1610\), \(z_3=\frac{1610 - 1800}{95}=\frac{-190}{95}=-2\)
For \(x = 1800\), \(z_4=\frac{1800 - 1800}{95}=0\)
Using technology (normalcdf\((-2,0)\))

For the third part, if we want the value \(x\) such that \(P(X>x)=0.09\), then \(P(X\leq x)=1 - 0.09 = 0.91\)
Using the inverse - normal function (e.g., invNorm\((0.91,1800,95)\))

Step2: Find probabilities and value

• For \(P(1705<X<1895)\):

Using the property of the normal distribution \(P(-1<Z<1)\approx0.6827\approx0.68\) (rounded to two decimal places)

• For \(P(1610<X<1800)\):

Using the property of the normal distribution \(P(-2<Z<0)\approx0.4772\approx0.48\) (rounded to two decimal places)

• For the value \(x\) where \(P(X>x) = 0.09\):

\(x=\text{invNorm}(0.91,1800,95)\approx1800+95\times1.34=1800 + 127.3=1927.30\)

Answer:

• The probability that a buyer paid between \(\$1705\) and \(\$1895\) is \(0.68\)
• The probability that a buyer paid between \(\$1610\) and \(\$1800\) is \(0.48\)
• The price a buyer would pay to get \(9\%\) the most expensive HD televisions is \(\$1927.30\)