Please help and solving with a calculator would be preferable! Thank you !

Question

The life of a radio shack player is normally distributed with a mean of 3.3 years and a standard deviation of 0.8 years. Radio shack guarantees its record players for 2 years. Find the probability that a record player will break down during the

Answer 1

When I first look at the problem, I want to guess about what the answer should be and then calculate the actual answer. If the two are compatible, I feel comfortable. If not, I think harder.

Guesswork:

1) Since the warranty time (2 years) is less than the mean lifetime and lifetime is normally distributed, I know the probability is less than 0.5. Why? Because, for a normally distributed variable, half the lifetimes are less than the mean and half are more. In other words, the population mean is the population median for normal distributions.

2) Since the warranty time is greater than 1.732 (mean - 1.96 * SD), the probability is greater than 0.025. Why? 2.5% of a normal distributed outcomes fall below 1.96 SD's below the mean and 2.5% are above 1.96 SD's above the mean. That's using a 95% confidence interval to figure out what the answer should be. The number 1.96 (or 2, if you prefer simplicity) is an easy number to remember for quick guesswork.

3) Since the warranty duration is much closer to 1.732 than to 3.3, the real answer should be a lot closer to 0.025 than to 0.5. Without calculating or thinking too long, I'd guess it's about 0.06. Why? Experience and it feels right ... nothing official.

Real answer:

Based on the assumptions (mean = 3.3, SD = 0.8, warranty duration of 2, normally distributed), I want to find the proportion of such lifetimes being less than 2 and I know the answer should be between 0.025 and 0.5, much closer to the small side. So, using an online calculator (I chose the Stat Trek Normal Distribution Calculator), I replace the entered values of 0 and 1 with my actual mean/SD of 3.3 and 0.8. Then, under "standard score", I enter the value of 2. I press calculate and get an answer of 0.052. That's the probability that the failure will occur anytime less than 2 years, also known as the cumulative probability of a value less than or equal to 2. Happily that fits well with my guesswork, so I can be pretty comfie with the answer.

Alternate version, in case a standard normal is required (not sure if this is still a thing):

Under the same assumptions as above, calculate how many SD's the interesting value (2) is below the mean. How? Compute: (value - mean)/SD. In this case, 2 - 3.3, is -1.3. Then divide that number by the SD (0.8) which gives -1.625. That means your interesting value is 1.625 standard deviations below the mean. Now calculate or look up how often a standard normal variable falls below -1.625 and ... tada ... same answer.

Answer 2

x = life

x is N(3.3,0.8)

assuming you are asking for the probability it will break down while under guarantee, then

you want to find P(x < 2)

transform inequality to standard normal (z), which is N(0,1).

P(z < (2-3.3)/0.8) = P(z < -2.03) = 0.0212

Answer 3

Since we are told this is a normal distribution (gaussian distribution), the parameters needed to determine the probability and fully quantify this function are the mean and std-deviation. With a mean of 3.3 years and a std-deviation of 0.8 years, we know the following...

1) 68% of samples fit within +/- 1 std-deviation of the mean : 2.5 ≤ p ≤ 4.1

2) 95% of samples fit within +/- 2 std-deviations of the mean : 1.7 ≤ p ≤ 4.9

3) 99.9% of samples fit within +/- 3 std-deviations of the mean : 0.9 ≤ p ≤ 5.7

So we have a 2 year warrant offering, so we can integrate from -∞ to 2 to get the probability of breakage of a unit...

1) int_{-∞}^{2} e^{-(x - μ)*(x - μ)/(2σ*σ)} dx = Probability of breaking before 2 years....

2) int_{2}^{∞} e^{-(x - μ)*(x - μ)/(2σ*σ)} dx = Probability of survival beyond 2 years....

https://onlinestatbook.com/2/calculators/normal_dist.html

Doing this integration for #1 yields (via an online calculator or a table) : 5.21%

Doing this integration for #2 yields (via an online calculator or a table) : 94.79%

So there is very good chance the unit will survive past 2 years with a low probability of breakage before then...