MATLAB 4. Assume that the values given below are sample values of the breaking s

Question

MATLAB

4. Assume that the values given below are sample values of the breaking strength in Newtons of a certain material, and it is known that breaking strengths are normally distributed. Plot the scaled frequency histogram using a bin width of 2. Compute the mean and standard deviation of the breaking strength sample values, and use them to estimate the lower and upper limits of breaking strength corresponding to 68.3 and 95.5 percent of such materials. Compare these limits with those obtained from the scaled frequency histogram. Using the sample estimates of the mean and standard deviation, compute the following probabilities for a randomly chosen material a. The breaking strength is greater than 19.5 Newtons. b. The breaking strength is between 20.2 and 24.5 Newtons. c. The breaking strength is no greater than 26.6 Newtons. Breaking Strengths (Newtons) 24.2122 17.6665 24.6686 21.7516 19.8550 20.9123 12.1747 19.5773 27.0197 20.4917 22.2069 16.3292 19.8459 21.1617 22.1319 18.1360 18.8177 30.1185 18.7082 32.1472 21.6878 27.0003 11.679217.0498 18.6097 22.1136 24.6490 22.3582 13.9358 20.3310 23.2618 21.6353 25.4132 21.0304 16.745521.2853 15.2781 13.3911 24.6241 20.000219.725424.5556 22.9729 18.7510 13.2513 24.6489 19.1988 16.5482 24.3086 20.0058 28.6458 7.5686 22.9059 19.0378

Explanation / Answer

Solution:

code:

F=[24.2122 12.1747 19.8459 18.7082 18.6097 23.2618 15.2781 22.9729 24.3086 17.6665 19.5773 21.1617 32.1472 22.1136 21.6353 13.3911 18.7510 20.0058 24.6686 27.0197 22.1319 21.6878 24.6490 25.4132 24.6241 13.2513 28.6458 21.7516 20.4917 18.1360 27.0003 22.3582 21.0304 20.002 24.6489 7.5686 19.8550 22.2069 18.8177 11.6792 13.9358 16.7455 19.7254 19.1988 22.9059 20.9123 16.3292 30.1185 17.0498 20.3310 21.2853 24.556 16.5482 19.0378];

x=5:35;

[n,xout]=hist(F,x);

binwidth=x(2)-x(1);
area=binwidth*sum(n);
mpg_scaled=n/area;
bar(x,mpg_scaled)
hold on
xlabel('Force')
ylabel('Scaled Frequency')
mu=mean(F)
sigma=std(F)
xplot=5:0.1:35;
p=1/(sigma*sqrt(2*pi))*exp(-(xplot-mu).^2/(2*sigma^2));
plot(xplot,p,'-r')

lower_limit=mu-sigma
upper_limit=mu+sigma
llplot=[lower_limit,lower_limit];
ulplot=[upper_limit,upper_limit];
yplot=[0 0.16];
plot(llplot,yplot,'-k',ulplot,yplot,'-k')
hold off

b1=19.5;
P1=1-(1/2)*(1+erf((b1-mu)/(sigma*sqrt(2))));
sprintf('probability of breaking strength greater than 19.5 Newtons ==> %f',P1)
a2=20.2;
b2=24.5;

P2=(1/2)*(erf((b2-mu)/(sigma*sqrt(2)))-erf((a2-mu)/(sigma*sqrt(2))));
sprintf('probability of breaking strength between 20.2 and 24.5 Newtons ==> %f',P2)
b3=26.6;
P3=(1/2)*(1+erf((b3-mu)/(sigma*sqrt(2))));
sprintf('probability of breaking strength less 26.6 Newtons ==> %f',P3)

I hope this helps if you find any problem. Please comment below. Don't forget to give a thumbs up if you liked it. :)

Related Questions

Navigate

• Browse (All)
• Browse M
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.