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?

(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:

*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:

*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:

*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:

*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:

*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:

*Use a z-table.

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

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step-by-step explanation: