Answer:
The probability that there will be at most 41 correct answers is 0.8790
Explanation:
We can aproximate the probability by using a binomial distribution where:
p: A question on a test is answered correctly
n: number of responses
So, the mean of the distribution is given by:
![\mu= n* p= 140 * 0.25=35](https://img.qammunity.org/2020/formulas/mathematics/high-school/qphbubkclby98vms0kj9fiqav8lhyxv5ej.png)
and the standar deviation is given by:
![\sigma=√(n* p* q) =√(n* p* (1-p)) =√(140* 0.25* 0.75) =5.123](https://img.qammunity.org/2020/formulas/mathematics/high-school/eylslgmauxm905093l2m3vphzod7tjk6b7.png)
The normalized variable for 41 correct answers is:
![z=(x-\mu)/(\sigma)=(41-35)/(5.123) =1.17](https://img.qammunity.org/2020/formulas/mathematics/high-school/muma8mx1nydtsnd1pgzsifwmihf0bym3ap.png)
Hence, the probability that there will be at most 41 correct answers is:
![P(x<41)=P(z<1.17)=0.8790](https://img.qammunity.org/2020/formulas/mathematics/high-school/s6mc47n0sps59l7zfabnessi8s4syd7ofp.png)