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Current research indicates that the distribution of the life expectancies of a certain protozoan is normal with a mean of 48 days and a standard deviation of 10.2 days. find the probability that a simple random sample of 49 protozoa will have a mean life expectancy of 49 or more days.

User Andykkt
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X is the random variable for lifespan of a protozoan and
X\sim\mathcal N(48,10.2^2). Let
\bar X be the mean of a sample from this distribution, so that
\bar X\sim\mathcal N\left(48,\left((10.2)/(√(49))\right)^2\right).

For the sake of clarity, I'm denoting a normal distribution with mean
\mu and standard deviation
\sigma by
\mathcal N(\mu,\sigma^2).

We have


P(\bar X\ge49)=P\left(\frac{\bar X-48}{(10.2)/(√(49))\ge\frac{49-48}{(10.2)/(√(49))\right)\approx P(Z\ge0.6863)\approx0.2463

(where
Z follows the standard normal distribution)

User Azmuhak
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