Mark is covering an 8-inch-square board with 1-inch-square tiles. He uses gray tiles to make four 2-in. by 2-in. squares. He cov
ers the rest of the board with white tiles.
How many more white tiles than gray tiles will Mark use?
Four small rectangles inside a larger rectangle. Two adjacent sides of the larger rectangle are labeled eight inches and eight inches. The four smaller rectangles appear to be the same size and are all shaded. Two adjacent sides of one of the smaller rectangles are labeled two inches and two inches.
How many more white tiles than gray tiles will Mark use?
Four small rectangles inside a larger rectangle. Two adjacent sides of the larger rectangle are labeled eight inches and eight inches. The four smaller rectangles appear to be the same size and are all shaded. Two adjacent sides of one of the smaller rectangles are labeled two inches and two inches.
1 answer:
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Answer:
(a) The distribution of is a Binomial distribution.
(b) The sampling distribution of the sample mean will be approximately normal.
(c) The value of is 0.50.
Step-by-step explanation:
It is provided that random variables are independent and identically distributed Bernoulli random variables with <em>p</em> = 0.50.
The random sample selected is of size, <em>n</em> = 100.
(a)
Theorem:
Let be independent Bernoulli random variables, each with parameter <em>p</em>, then the sum of of thee random variables,
is a Binomial random variable with parameter <em>n</em> and <em>p</em>.
Thus, the distribution of is a Binomial distribution.
(b)
According to the Central Limit Theorem if we have an unknown population with mean <em>μ</em> and standard deviation <em>σ</em> and appropriately huge random samples (<em>n</em> > 30) are selected from the population with replacement, then the distribution of the sample mean will be approximately normally distributed.
The sample size is large, i.e. <em>n</em> = 100 > 30.
So, the sampling distribution of the sample mean will be approximately normal.
The mean of the distribution of sample mean is given by,
And the standard deviation of the distribution of sample mean is given by,
(c)
Compute the value of as follows:
*Use a <em>z</em>-table.
Thus, the value of is 0.50.