Question

Solve the problem. A ompany manufactures batteries in batches of 6 and there is a 3% rate of defects. Find the mean number of dele per batch. D) 18 9 02 A) 5.8 B) 18 Estimate the indicated probability by using the normal distribution as an approximation to the binomial distribution. 32) The probability that a radish seed will germinate is 0.7, Eatimate the probability that of 140 randomly selected seeds, exactly 100 will germinate. A) 0.0679 B) 0.9331 C) 0.0669 D) 0.0769 Find the indicated probability. 33) A bank's loan officer rates applicants for credit. The ratings are normally distributed with a mean of 200 and a standard deviation of 50. If an applicant is randomly selected, find the probability of a rating that is between 200 and 275. A) 04332 B) 0.9332 C) 0.0668 D) 0.5 If z is a standard normal variable, find the probability. 34) P(z <0.97) A) 0.8078 B) 0.8340 C) 0.1660 D) 0.8315 If z is a standard normal variable, find the probability. 35) P(-0.73 <z< 2.27) A) 0.7557 B) 1.54 ) 0.4884 D) 0.2211 Find the area of the shaded region. The graph depicts the standard normal distribution with mean 0 and standard deviation 1. 36) adi b o sulev ba dw on 3.39 2.26 -L.13 L13 2.26 3.39 A) 0.1292 B) 0.8708 C) 0.8485 D) 0.8907

Answer 1

As per HomeworkLib policy i answered the first 5 questions

31) Answer: ----- Given, Date: ----12/12/2019 n=6 and p=0.03 Using Binomial distribution, Mean, pl = n*p=6*0.03 = 0.18 Therefore, mean number of defects is 0.18 option---D

32) Answer: ----

Probability that a radish seed will germinate, p=0.7 Let X denote number of seeds germinate from a sample of 140. Here, X ~ Binom(n=140, p = 0.7) Calculate the probability that exactly 100 seeds will germinate. Approximation: To find the above probability use normal approximation because both np and n(1- p) are greater than 5. Mean, u = np = 140x0.7=98 Standard deviation, o = vnp(1- p) = 140(0.7)(1 – 0.7) = 5.422177 Now, determine the required probability. P( X = 100)=P(100 -0.5< X <100+ 0.5) = P(99.5 < X <100.5) (99.5 - 98 X - 100.5-98 15.422177. o 5.422177) = P(0.28< Z<0.46) = P(Z <0.46) - P(Z <0.28) = NORMSDIST (0.46) - NORMSDIST (0.28) = 0.6772 -0.6103 {Use Excel function or z-table = 0.0669 Required answer: 0.0669.

Exact probability: The PMF of X is, P(X = x) =(. \p*(1- p)** 0.3)**, x = 0,1,2,...,140 Now, calculate the probability that exactly 100 seeds will germinate. P(X=100)=(140 (0.7)100 (0.3)760-100 100 = 0.06952091 0.0695 Option----C

33) Answer: ---- The probability that the rating is between 200 and 275 is obtained as shown below: From the given information, let the random variable X be the ratings follows normal distribution with population mean u as 200 and population standard deviation o as 50. That is, j = 200,0 = 50 The required probability is, P (200 < X < 275) = P (2005-200 < 174 <275 50 = P (60 <<< = P(0.00 < 3 <1.50) = P(Z < 1.50) - P(Z <0.00) From the "standard normal table", the area to the left of Z = 1.50 is, Procedure for finding the probability value (confidence level) from standard normal table: The z value be 1.50. 1.In the standard normal table first locate the value 1.50 in the first z column. 2.Locate the value of 1.50 in the first z row. 3.Move right until the column of 0.00 is reached. 4.From the located 0.00 column move down until the row 1.50 is reached. 5.Locate the probability value, by the intersection of the row and column values gives the area to the left of z. The area to the left of Z = 1.50 is 0.9332. From the standard normal table", the area to the left of Z = 0.00 is, Procedure for finding the probability value (confidence level) from standard normal table: The z value be 0.00. 6.In the standard normal table first locate the value 0.00 in the first z column. 7. Locate the value of 0.00 in the first z row.

8. Move right until the column of 0.00 is reached. 9.From the located 0.00 column move down until the row 0.00 is reached. 10.Locate the probability value, by the intersection of the row and column values gives the area to the left of z. The area to the left of Z = 0.00 is 0.0500. P (200<x<275) = P(Z < 1.50) - P(Z <0.00) = 0.9332 – 0.500 = 0.4332 option-A The probability that the rating is between 200 and 275 is 0.4332.

34) Answer: -- P(z<0.97)=no msdist(0.97)=0.833976=0.8340 Option---B 35) Answer: --- P(-0.73 <<2.27)=0.9884-0 2327=0.7557 Option--A