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The amount of time a certain brand of light bulb lasts is normally distributed with a mean of $2000$ hours and a standard deviation of $65$ hours. Out of $590$ freshly installed light bulbs in a new large building, how many would be expected to last less than $1850$ hours, to the nearest whole number?

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Interpret confidence intervals for population means

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Q. The amount of time a certain brand of light bulb lasts is normally distributed with a mean of

2000

hours and a standard deviation of

65

hours. Out of

590

freshly installed light bulbs in a new large building, how many would be expected to last less than

1850

hours, to the nearest whole number?

Calculate z-score: Calculate the z-score for $1850$ hours using the formula $z = \frac{X - \mu}{\sigma}$ , where $X$ is the value in question, $\mu$ is the mean, and $\sigma$ is the standard deviation. $\newline$ $X = 1850$ hours, $\mu = 2000$ hours, $\sigma = 65$ hours. $\newline$ $z = \frac{1850 - 2000}{65}$ $\newline$ $z = \frac{-150}{65}$ $\newline$ $z = \frac{X - \mu}{\sigma}$ $0$
Lookup in table: Look up the z-score in a standard normal distribution table to find the probability that a light bulb will last less than $1850$ hours. $\newline$ For $z \approx -2.31$ , the probability (p-value) is approximately $0.0107$ .
Calculate expected number: Calculate the expected number of light bulbs that will last less than $1850$ hours by multiplying the total number of light bulbs by the probability found from the z-score. $\newline$ Total number of light bulbs $= 590$ $\newline$ Expected number $= 590 \times 0.0107$ $\newline$ Expected number $\approx 6.313$
Round to nearest whole: Round the expected number to the nearest whole number since we cannot have a fraction of a light bulb. $\newline$ Expected number $\approx 6$ (to the nearest whole number)

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