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)

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belgrad belgrad · Answer 1 · 2021-04-23T05:54:29+02:00

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

[tex]z_{1} = \frac{x_{1}-mean }{S.D} = \frac{800-850}{50} = -1[/tex]

Let x₂ =850

[tex]z_{2} = \frac{x_{2}-mean }{S.D} = \frac{900-850}{50} = 1[/tex]

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