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.
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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.
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Answer:
B. No, this distribution does not appear to be normal
Step-by-step explanation:
Hello!
To observe what shape the data takes, it is best to make a graph. For me, the best type of graph is a histogram.
The first step to take is to calculate the classmark`for each of the given temperature intervals. Each class mark will be the midpoint of each bar.
As you can see in the graphic (2nd attachment) there are no values of frequency for the interval [40-44] and the rest of the data show asymmetry skewed to the left. Just because one of the intervals doesn't have an observed frequency is enough to say that these values do not meet the requirements to have a normal distribution.
The answer is B.
I hope it helps!
