A manufacturer knows that their items have a normally distributed lifespan, with a mean of 8.8 years, and standard deviation of 2.2 years.

If 5 items are picked at random, 5% of the time their mean life will be less than how many years?

First of all we need o use the standard normal table to see what value if z will have 10% of the area (probability) to the left of it. We see that z value is approximately z = -1.282

Now all we have to do is calculate the value of x that corresponds to this value of z. We have

( x - µ ) / σ = -1.282 Substituting in the known values, we have

( x - 4.4 ) / 1.3 = -1.282 and x = 2.73 years

A manufacturer knows that their items have a normally distributed lifespan, with a mean of 8.8 years, and standard deviation of 2.2 years. If 5 items are picked at random, 5% of the time their mean life will be less than how many years?

Answer :

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A manufacturer knows that their items have a normally distributed lifespan, with a mean of 8.8 years, and standard deviation of 2.2 years.

If 5 items are picked at random, 5% of the time their mean life will be less than how many years?