Question

Suppose that lithium calculator battery with a charge of 3 volts, actually has a voltage with uniform distribution between 2.93 and 3.1 volts. What are the expected voltage and standard deviation of voltage? What is the probability that a battery has a voltage less than 2.98?
a. E(X) = 2.413, STDEV(X)= 0.0024, P (X ≤ 2.98) = 0.3342
b. E(X) = 3.015, STDEV(X)= 0.049, P (X ≤ 2.98) = 0.2941
c. E(X) = 3.015, STDEV(X)= 0.0049, P (X ≤ 2.98) = 0.1041
d. None of the preceding
e. E(X) = 3.015, STDEV(X)= 0.0024; P (X ≤ 2.98) = 0.2941
f. E(X) = 2.413, STDEV(X)= 0.049, P (X ≤ 2.98) = 0.2941

Answer 1

Answer:

b. E(X) = 3.015, STDEV(X)= 0.049, P (X ≤ 2.98) = 0.2941

Step-by-step explanation:

An uniform probability is a case of probability in which each outcome is equally as likely.

For this situation, we have a lower limit of the distribution that we call a and an upper limit that we call b.

The mean of the uniform probability distribution is:

$M = \frac{a + b}{2}$

The standard deviation of the uniform distribution is:

$S = \sqrt{\frac{(b-a)^{2}}{12}}$

The probability that we find a value X lower than x is given by the following formula.

$P(X \leq x) = \frac{x - a}{b-a}$

Uniform distribution between 2.93 and 3.1 volts

This means that $a = 2.93, b = 3.1$. So

Mean:

$M = \frac{2.93 + 3.1}{2} = 3.015$

Standard deviation:

$S = \sqrt{\frac{(3.1 - 2.93)^{2}}{12}} = 0.049$

What is the probability that a battery has a voltage less than 2.98?

$P(X \leq 2.98) = \frac{2.98 - 2.93}{3.1 - 2.93} = 0.2941$

So the correct answer is:

b. E(X) = 3.015, STDEV(X)= 0.049, P (X ≤ 2.98) = 0.2941