The letter "t" makes up an estimated 10% of a certain language. Assume this is still correct. A random sample of 500 letters is taken from a randomly selected, large book and the t's are counted. Find the approximate probability that the random sample of 500 letters will contain 9.1% or fewer t's.

  0.2755

Step-by-step explanation:

We intend to make use of the normal approximation to the binomial distribution.

First we'll check to see if that approximation is applicable.

For p=10% and sample size n = 500, we have ...

  pn = 0.10(500) = 50

This value is greater than 5, so the approximation is valid.

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The mean of the distribution we'll use as a model is ...

  µ = p·n = 0.10(500)

  µ = 50

The standard deviation for our model is ...

  σ = √((1-p)µ) = √(0.9·50) = √45

  σ ≈ 6.708204

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A continuity correction can be applied to better approximate the binomial distribution. We want p(t ≤ 9.1%) = p(t ≤ 45.5). For our lookup, we will add 0.5 to this limit, and find p(t ≤ 46).

The attached calculator shows the probability of fewer than 45.5 t's in the sample is about 0.2755.