Question

A regular reader of BYTE magazine, Fritz, has decided to purchase a computer with these particular specifications. These computers are have an approximate normal distribution with a mean of $850 and a standard deviation of $50. His budget is flexible, and he can afford to spend between $800 and $900. What percentage of these computers are within his budget? (Use the Empirical Rule 68, 95, 99.7)

Answer 1

Answer:

The probability that can afford to spend between $800 and $900

P(800≤X≤900) = 0.6826

The percentage of that can afford to spend between $800 and $900

P(800≤X≤900) = 68 percentage

Step-by-step explanation:

Step(i):-

Given that the mean of the Normal distribution = $850

Given that the standard deviation of the Normal distribution =  $50

Let 'X' be a random variable in a normal distribution

Let x₁ = 800

$z_{1} = \frac{x_{1}-mean }{S.D} = \frac{800-850}{50} = -1$

Let x₂ =850

$z_{2} = \frac{x_{2}-mean }{S.D} = \frac{900-850}{50} = 1$

Step(ii):-

The probability that can afford to spend between $800 and $900

P(800≤X≤900) = P(-1≤Z≤1)

                         = P(Z≤1) - P(Z≤-1)

                        = 0.5 + A(1) - (0.5 - A(-1))

                        = A(1) +A(-1)

                       = 2× A(1)       (∵ A(-1) =A(1)

                      = 2 × 0.3413

                      = 0.6826

The percentage of that can afford to spend between $800 and $900

P(800≤X≤900) = 68 percentage