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vladimir2022 [97]
3 years ago
10

The letter​ "t" makes up an estimated 10​% of a certain language. Assume this is still correct. A random sample of 500 letters i

s 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.

Mathematics
1 answer:
Andrew [12]3 years ago
6 0

Answer:

  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.

__

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

__

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.

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Step-by-step explanation:

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3 years ago
The distribution of annual profit at a chain of stores was approximately normal with mean \mu = \$66{,}000μ=$66,000mu, equals, d
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Answer:

The closest to the maximum profit is  x = \$ 83682

Step-by-step explanation:

From the question we are told that

  The  mean is  \mu  =  \$66,000

   The standard deviation is  \sigma  = \$ 21000

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Generally the closest to the maximum annual profit at a store where the executives conducted an audit is mathematically evaluated  as follows

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=>  P(X >  x ) = P(\frac{X -x}{\sigma }  >  \frac{x -66000}{21000} ) =  0.20

From the z-table  the z-score for  0.20  is  

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4 years ago
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