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Check each binomial distribution to see whether it can be approximated by a normal distribution (i.e., are $n p \geq 5$ and $n q \geq 5 ?$ ).a. $n=50, p=0.2$b. $n=30, p=0.8$c. $n=20, p=0.85$

Intro Stats / AP Statistics

Chapter 6

The Normal Distribution

Section 4

The Normal Approximation to the Binomial Distribution

Probability Topics

Temple University

University of North Carolina at Chapel Hill

Piedmont College

Lectures

0:00

02:42

Check each binomial distri…

08:23

Recall that for use of a n…

14:39

Use the normal approximati…

08:31

04:39

In which of the following …

02:19

Use a computer to find the…

04:40

a. Use a computer or calcu…

02:29

Find the normal approximat…

04:25

Find the mean, variance, a…

02:43

04:56

Consider the binomial expe…

04:41

Let $x$ be a binomial rand…

01:32

Probability Distribution A…

01:36

Suppose that $x$ has a bin…

02:48

In order to see what happe…

04:35

01:43

Find the values $n p$ and …

01:15

02:57

a. Use a calculator or com…

03:59

Consider a binomial experi…

Okay, So for problem number eight, there's, too was two stipulations in order to use a normal distribution as an estimate for a binomial distribution. Both of these need to be satisfied. End Times P is greater than or equal to five and n times. Q is also greater than or equal to five. Okay, we need both of these to be true. So for part A for NPR, we have 50 times 0.2, and that's going to give us 10. And for N Q. We have 50 times one minus point to okay, Cue is just one minus p. So it's one minus point to, and that's going to give us 40. Okay, Both of these answers are greater than or equal to five. So Port A is yes, part B. For NPR, we have 30 times 0.8, which gives us 24. And for N Q. We have 30 times one minus 10.8, which gives us six. Okay, now 24 is greater than or equal to five, and six is greater than or equal to five. Since we satisfied both conditions, the answer is yes. For port, See and P, we have 20 times 0.85 and that gives of 17. And for N Q. We have 20 times one minus 10.85 and that gives us three. Okay, 17 is greater than or equal to five, but three is not okay. So the answer for C is no. All right again. You need to satisfy both conditions, or you can't use the normal distribution as an approximation.

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