Question

The graph illustrates a normal distribution for the prices paid for a particular model of HD television. The mean price paid is $1600 and the standard deviation is $100. What is the approximate percentage of buyers who paid more than $1900? What is the approximate percentage of buyers who paid less than $1400?What is the approximate percentage of buyers who paid between $1400 and $1600?What is the approximate percentage of buyers who paid between $1500 and $1700?What is the approximate percentage of buyers who paid between $1600 and $1700?What is the approximate percentage of buyers who paid between $1600 and $1900?

Answer 1

Answer:

(a) 0.14%

(b) 2.28%

(c) 48%

(d) 68%

(e) 34%

(f) 50%

Step-by-step explanation:

Let X be a random variable representing the prices paid for a particular model of HD television.

It is provided that X follows a normal distribution with mean, μ = $1600 and standard deviation, σ = $100.

(a)

Compute the probability of buyers who paid more than $1900 as follows:

$P(X>1900)=P(\frac{X-\mu}{\sigma}>\frac{1900-1600}{100})$

                   $=P(Z>3)\\=1-P(Z<3)\\=1-0.99865\\=0.00135\\\approx 0.0014$

*Use a z-table.

Thus, the approximate percentage of buyers who paid more than $1900 is 0.14%.

(b)

Compute the probability of buyers who paid less than $1400 as follows:

$P(X<1400)=P(\frac{X-\mu}{\sigma}<\frac{1400-1600}{100})$

                   $=P(Z<-2)\\=1-P(Z<2)\\=1-0.97725\\=0.02275\\\approx 0.0228$

*Use a z-table.

Thus, the approximate percentage of buyers who paid less than $1400 is 2.28%.

(c)

Compute the probability of buyers who paid between $1400 and $1600 as follows:

$P(1400<X<1600)=P(\frac{1400-1600}{100}<\frac{X-\mu}{\sigma}<\frac{1600-1600}{100})$

                              $=P(-2<Z<0)\\=P(Z<0)-P(Z<-2)\\=0.50-0.0228\\=0.4772\\\approx 0.48$

*Use a z-table.

Thus, the approximate percentage of buyers who paid between $1400 and $1600 is 48%.

(d)

Compute the probability of buyers who paid between $1500 and $1700 as follows:

$P(1500<X<1700)=P(\frac{1500-1600}{100}<\frac{X-\mu}{\sigma}<\frac{1700-1600}{100})$

                              $=P(-1<Z<1)\\=P(Z<1)-P(Z<-1)\\=0.84134-0.15866\\=0.68268\\\approx 0.68$

*Use a z-table.

Thus, the approximate percentage of buyers who paid between $1500 and $1700 is 68%.

(e)

Compute the probability of buyers who paid between $1600 and $1700 as follows:

$P(1600<X<1700)=P(\frac{1600-1600}{100}<\frac{X-\mu}{\sigma}<\frac{1700-1600}{100})$

                              $=P(0<Z<1)\\=P(Z<1)-P(Z<0)\\=0.84134-0.50\\=0.34134\\\approx 0.34$

*Use a z-table.

Thus, the approximate percentage of buyers who paid between $1600 and $1700 is 34%.

(f)

Compute the probability of buyers who paid between $1600 and $1900 as follows:

$P(1600<X<1900)=P(\frac{1600-1600}{100}<\frac{X-\mu}{\sigma}<\frac{1900-1600}{100})$

                              $=P(0<Z<3)\\=P(Z<3)-P(Z<0)\\=0.99865-0.50\\=0.49865\\\approx 0.50$

*Use a z-table.

Thus, the approximate percentage of buyers who paid between $1600 and $1900 is 50%.