In a family with children, the probability that all the children are girls is appoximately . In a random sample of 1000 families with children, what is the approximate probability that or fewer will have girls? Approximate a binomial distribution with a normal distribution.

The probability that 100 or fewer will have 3 girls is 0.00734.

Step-by-step explanation:

The complete question is:

In a family with 3 children, the probability that all the children are girls is approximately 0.125. In a random sample of 1000 families with 3 children, what is the approximate probability that 100 or fewer will have 3 girls? Approximate a binomial distribution with a normal distribution.

Solution:

Let X represent the number of families who has 3 girls.

The random variable X follows a Binomial distribution with parameters n = 1000 and p = 0.125.

But the sample selected is too large.

So a Normal approximation to binomial can be applied to approximate the distribution of X if the following conditions are satisfied:

1.np ≥ 10

2.n(1 - p) ≥ 10

Check the conditions as follows:

Thus, a Normal approximation to binomial can be applied.

So,  

The mean and standard deviation are:

Compute the probability that 100 or fewer will have 3 girls as follows:

Apply Continuity correction:

*Use a z-table.

Thus, the probability that 100 or fewer will have 3 girls is 0.00734.

y = 3x+4

step-by-step explanation:

to find the slope

slope = (y2-y1)/ (x2-x1)

    = (-8-4)/ (-4-0)

    = -12/-4

  = 3

using the slope intercept form of a line

y= mx+b

we know the slope is 3

we know the y intercept is 4 ( the point where it crosses the y axis or x=0)

y = 3x+4

for the free

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