A dart board contains 20 equally-sized sectors numbered 1 to 20. A dart is rando

Cheyanne Leigh

Cheyanne Leigh

Answered question

2021-09-28

Compute P(X) using the binomial probability formula. Then determine whether the normal distriution can be used to estimate this probability. If so, approximate P(X) using the normal distriution and compare the result with exact probability.
n=47,p=0.6, and X=34
For n=47,p=0.6, and X=34, use the binomial probability formula to find P(X).
(Round to four decimal plases as needed.)
Can be normal distriution be used to approximate this probability?
A. Yes, because np(1p)10
B. No, because np(1p)10
C. No, because np(1p)10
D. Yes, because np(1p)10

Answer & Explanation

Jaylen Fountain

Jaylen Fountain

Skilled2021-09-29Added 169 answers

Step 1
Obtain the value of P(X) using the binomial Probability formula.
The value of P(X) using the binomial Probability formula is obtained below as follows:
Let X denotes the random variable which follows binomial distribution with the probability of success 0.60 and the number of trails selected is 47. That is, XB(47,0.60)
That is n=47,p=0.60,q=0.40(=10.60).
The probability distribution is given by,
P(X=x)=(nx)px(1p)nx; here x=0,1,2,,n for 0p1
Where n is the number of tials and p is the probability of success for each trial
The required probability is,
P(X=34)=(4737)(0.60)34(10.60)4734
=(4737)(0.60)34(0.40)13
=0.0270
The value of P(X) using the binomial Probability formula is 0.0270.
Step 2
Determine whether the normal distribution can be used to approximate this probability.
The normal distribution can be used to approximate this probability or not determined below as follows:
That is,
np(1p)=47×0.60×(10.60)
=47×0.60×0.40
=11.28
10
Since, the value of greater than or equal 10, it can be concluded that normal distribution can be used to approximate this probability.
Correct option: Option D
Step 3
Obtain the value of mean.
The value of mean is obtained below as follows:
E(X)=np
=47×0.60
=28.2
Obtain the value of standard deviation.
The value of standard deviation is obtained below as follows:
SD(X)=npq
=47×0.60×0.40
=11.28
=3.3586
Obtain the approximated probability of X=34.
The approximated probabllity of X=34 is obtained below as follows:
The required probability is,
P(X=34)=P(340.5X34+0.5)
=P(33.5X34.5)
=P(33.528.23.3586Z34.528.23.3586)
=P(5.33.3586Z6.33.3586)
=P(1.58Z1.88)
K=P(Z1.88)P(Z1.58)[FromEXCELNORMSDIST(1.88)NORMSDIST(1.58)]
=0.96990.9429
=0.0270
The approximated probability of X=34 is 0.0270.
Compare the result with the exact probability.
difference =0.02700.0270
0
The result with the exact probability is 0.

Do you have a similar question?

Recalculate according to your conditions!

New Questions in High school probability

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?