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.
1 answer:
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.
You might be interested in
Answer:
A
Step-by-step explanation:
We are given that:
And we want to find:
Remember that tangent and cotangent are co-functions. In other words, they follow the cofunction identities:
Therefore, since tan(θ) = 1.3 and cot(90° - θ) = tan(θ), then cot(90° - θ) must also be 1.3.
Our answer is A.