6.0. Help -1 If op2 5 and nq 2 5, estimate P(at least 3) with n 13 and p 05 by using the normal distribution as an approximation to the binomial distribution, if np<5orng 5 then state that the normal approximation is not suitable Select the correct choice below and, if necessary, fill in Ithe answer box to complete your choice A. P(at least 3) (Round to three decimal places as needed) B. The nomal distribution cannot be used Click to select and enter your answer(s) and then click Check Answer Check Ansaet Clear All All parts showing P Pearson 937 PA AW&EAM11/2/2019 e search

MATLAB: An Introduction with Applications
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ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
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-1
If op2 5 and nq 2 5, estimate P(at least 3) with n 13 and p 05 by using the normal distribution as an approximation to the binomial distribution, if np<5orng 5
then state that the normal approximation is not suitable
Select the correct choice below and, if necessary, fill in Ithe answer box to complete your choice
A. P(at least 3)
(Round to three decimal places as needed)
B. The nomal distribution cannot be used
Click to select and enter your answer(s) and then click Check Answer
Check Ansaet
Clear All
All parts showing
P Pearson
937 PA
AW&EAM11/2/2019
e
search
Transcribed Image Text:6.0. Help -1 If op2 5 and nq 2 5, estimate P(at least 3) with n 13 and p 05 by using the normal distribution as an approximation to the binomial distribution, if np<5orng 5 then state that the normal approximation is not suitable Select the correct choice below and, if necessary, fill in Ithe answer box to complete your choice A. P(at least 3) (Round to three decimal places as needed) B. The nomal distribution cannot be used Click to select and enter your answer(s) and then click Check Answer Check Ansaet Clear All All parts showing P Pearson 937 PA AW&EAM11/2/2019 e search
Expert Solution
Step 1

Given data

n = 13

p = 0.5

np = 13 x 0.5 = 6.5 > 5

nq = 13 x (1 – 0.5 ) = 6.5 > 5

So, conditions for the normal distribution as an approximation to the binomial distribution is valid.

Mean µ = np = 6.5

Standard Deviation is given by 

1.803
прд — V13 х0.5 х0.5
P(At least 3) is given by
Р(X> 3)
Changing into standard normal variate
3 6.5
X
= P(z 2-1.941)
= Pz 2
P
1.803
1 - P(z < 1.941) = 1 - 0.974 -
P(z
1.941)
0.026
(P (z 1.941) = 0.974(From Excel NORM. S. DIST(1.941, TRUE))
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