According to a survey, 61% o murders committed last year were cleared by arrest

Question

According to a survey, 61% o murders committed last year were cleared by arrest or excep onal means Fifty murders comm ed last year are an ons y selec o the number cleared by arrest or exceptional means is recorded When technology is used, use the Tech Help button for further assistance , and t up (a) Find the probability that exactly 40 of the murders were cleared (b) Find the probability that between 36 and 38 of the muirders, inclusive, were cleared fx() Woukd it be unusual if fewer than 19 of the murders were cleared? Why or why not? a) The probability that exactly 40 of the murders were cleared is Round to four decimal places as needed) b) The probability that between 36 and 38 of the murders, inclusive, were cleared is Round to four decimal places as needed) (c) Would it be unusual if fewer than 19 of the murders were cleared? Why or why not? O A. Yes it would be unusual because 19 s less than -20 B. Yes, it would be unusual because 19 between -2 and + 2a SEL N C. D. Not would not be unusual because 19 is less than -20 No, it would not be unusual because 19 is between -20 and-2n Click to select your answer(s) LEARNIN Online T Chegg Sh STUDENTISERVICES GOURCES e Map obne Retun, Your Books Textbook oner Servic

Explanation / Answer

the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
mean ( np ) = 50 * 0.61 = 30.5
standard deviation ( npq )= 50*0.61*0.39 = 3.4489
equation of the normal curve is ( Z )= x - u / sd/sqrt(n) ~ N(0,1)
a)
P(X=40), assuming with sample error we calculate for error = 0.05 ,
P(X=40) = P(39.5 < X < 40.5)
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 39.5) = (39.5-30.5)/3.4489
= 9/3.4489 = 2.6095
= P ( Z <2.6095) From Standard Normal Table
= 0.99547
P(X < 40.5) = (40.5-30.5)/3.4489
= 10/3.4489 = 2.8995
= P ( Z <2.8995) From Standard Normal Table
= 0.99813
P(39.5 < X < 40.5) = 0.99813-0.99547 = 0.0027
b)
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 36) = (36-30.5)/3.4489
= 5.5/3.4489 = 1.5947
= P ( Z <1.5947) From Standard Normal Table
= 0.94461
P(X < 38) = (38-30.5)/3.4489
= 7.5/3.4489 = 2.1746
= P ( Z <2.1746) From Standard Normal Table
= 0.98517
P(36 < X < 38) = 0.98517-0.94461 = 0.0406
c)
P(X < 19) = (19-30.5)/3.4489
= -11.5/3.4489= -3.3344
= P ( Z <-3.3344) From Standard NOrmal Table
= 0.0004
optiona A: Yes, it would be unusual because 19 is less than u=2sigma

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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