Question

the amount of time a certain brand of light bulb lasts is normally distributed with a mean of 1200 hours and a standard deviation of 60 hours. what percentage light bulbs last less than 1160 hours, to the nearest tenth? statistics calculator

Explanation:

Step1: Calculate the z - score

The formula for the z - score is $z=\frac{x-\mu}{\sigma}$, where $x$ is the value from the dataset, $\mu$ is the mean, and $\sigma$ is the standard deviation.
Given that $\mu = 1200$, $\sigma=60$, and $x = 1160$.
Substitute these values into the formula: $z=\frac{1160 - 1200}{60}=\frac{- 40}{60}\approx - 0.67$ (rounded to two decimal places).

Step2: Find the cumulative probability

We need to find $P(Z < - 0.67)$, where $Z$ is a standard normal random variable.
Using a standard normal distribution table or a calculator, we know that the cumulative distribution function for the standard normal distribution gives the probability that $Z$ is less than a given value.
Looking up $z=-0.67$ in the standard normal table or using a calculator (for example, the normalcdf function on a TI - 84 plus: normalcdf(-1000, - 0.67,0,1)), we find that $P(Z < - 0.67)\approx0.2514$.

Step3: Convert to percentage

To convert the probability to a percentage, we multiply by 100: $0.2514\times100 = 25.14\%$.
Rounding to the nearest tenth, we get $25.1\%$.

Answer:

$25.1\%$