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A phone needs a battery life of at least 5 hours to be considered sufficient. You are in the process of buying a used phone, and notice that the battery life for each used phone (without knowing any other details than it being used) can be modelled as a normal distribution with mean of 4 hours, and a standard deviation of 1 hour. You decide to purchase 5 phones randomly. What are the odds that you get exactly 2 sufficient phone batteries?
Check your answer
Solution
Answer: 0.15
A phone is considered sufficient if the battery lasts at least \(5\) hours.
Thus we first find
\[
P(X \ge 5).
\]
\[
Z = \frac{X-\mu}{\sigma}.
\]
For \(X=5\),
\[
Z = \frac{5-4}{1} = 1.
\]
So this means that:
\[
P(X \ge 5) = P(Z \ge 1).
\]
Using the standard normal distribution,
\[
P(Z \le 1) \approx 0.84.
\]
Thus:
\[
P(Z \ge 1) = 1 - 0.8413 = 0.1587.
\]
So the probability that a single phone has a sufficient battery is
\[
p = 0.1587.
\]
Now suppose \(5\) phones are purchased independently.
Let \(Y\) be the number of sufficient batteries:
\[
Y \sim B(5,0.1587).
\]
We want
\[
P(Y=2).
\]
Using the binomial formula,
\[
P(Y=2) = \binom{5}{2}(0.1587)^2(1-0.1587)^3.
\]
\[
P(Y=2) \approx 0.150.
\]
So the probability of getting exactly \(2\) sufficient phone batteries out of \(5\) is approximately 0.15, or about \(15\%\).