Assume that there is a 6​% rate of disk drive failure in a year. a. If all your computer data is stored on a hard disk drive with a copy stored on a second hard disk​ drive, what is the probability that during a​ year, you can avoid catastrophe with at least one working​ drive? b. If copies of all your computer data are stored on three three independent hard disk​ drives, what is the probability that during a​ year, you can avoid catastrophe with at least one working​ drive?

Answer:

The mean of the distribution is 4.00 and the standard deviation is 2.31

Explanation:

Given

Range = 2 to 10

Type of distribution = uniform distribution

Required

1. Mean

2. Standard Deviation

The mean of uniform distribution is calculated as thus;

[tex]Mean = \frac{1}{2}(a + b)[/tex]

Where b and a are the intervals of the distribution

b = upper bound = 10

a = lower bound = 2

So,

[tex]Mean = \frac{1}{2}(a + b)[/tex]

Substitute 10 for b and 2 for a

[tex]Mean = \frac{1}{2}(2 + 10)[/tex]

[tex]Mean = \frac{1}{2}(12)[/tex]

[tex]Mean = \frac{1}{2} * 12[/tex]

[tex]Mean = 6[/tex]

[tex]Mean = 6.00[/tex] (Approximated to 2 decimal places)

The standard deviation of uniform distribution is calculated as thus;

σ = √σ²

Where σ represents the standard deviation and σ² represents the variance.

Calculate variance

σ² = Var

[tex]Var = \frac{(b-a)^2}{12}[/tex]

Substitute 10 for b and 2 for a

[tex]Var = \frac{(10 - 2)^2}{12}[/tex]

[tex]Var = \frac{8^2}{12}[/tex]

[tex]Var = \frac{64}{12}[/tex]

[tex]Var = 5.33[/tex]

Recall that

σ = √σ² = √Var

Substitute 5.33 for Var

σ = √5.33

σ = 2.309401076758503

σ = 2.31 (Approximated)

Hence, the mean of the distribution is 4.00 and the standard deviation is 2.31