Questions and Answers #1 Binomial Probability

Questions and Answers Sheet 1 Binomial Probability Question #1 If you buy a lottery ticket in 50 lotteries, in each of which your chance of winning a prize is 1 100 , what is the (approximate) probability that you will win a prize (a) at least once, (b) exactly once, and (c) at least twice? Answer: Step 1 𝑃{𝑋 = 𝑖} ≈ 𝑒 −𝜆 𝜆𝑖 𝑖! Given: 𝑛 = 50 1 𝑝= = 0.01 100 So, 𝜆 = 𝑛𝑝 = 50 × 0.01 = 0.5 a. Probability that you will win a prize at least once 𝑃{𝑋 ≥ 1} ≈ 1 − 𝑃{𝑋 = 0} = 1 − 𝑒 −0.5 0.50 0! = 0.3935 b. Probability that you will win a prize exactly once 𝑃{𝑋 = 0} ≈ 𝑒 −0.5 0.51 1! = 0.3033 c. Probability that you will win a prize at least twice 𝑃{𝑋 ≥ 2} ≈ 1 − 𝑃{𝑋 = 0} − 𝑃{𝑋 = 1} = 1 − 𝑒 −0.5 0.50 0! − 𝑒 −0.5 0.51 1! Question #2 Consider a binomial experiment with 𝑛 = 20 and 𝑝 = 0.70 a) Compute 𝑓(12) b) Compute 𝑓(16) c) Compute 𝑃(𝑥 ≥ 16) d) Compute 𝑃(𝑥 ≤ 15) e) Compute 𝐸(𝑥) f) Compute var(x) and \sigma Answer: Step 1 Given: 𝑝 = 0.70 𝑛 = 20 Formula binomial probability: n f (k ) = ( )  p k  (1 − p)n−k k a 20 f (12) = ( )  0.7012  (1 − 0.70)20−12 = 0.114397 12 b 20 f (16) = ( )  0.7016  (1 − 0.70)20−16 = 0.130421 16 c) Add the corresponding probabilities: 𝑃(𝑋 ≥ 16) = 𝑓(16) + 𝑓(17) + 𝑓(18) + 𝑓(19) + 𝑓(20) = 0.2375 = 0.0902 d) Use the complement rule for probabilities: 𝑃(𝑋 ≤ 15) = 1 − 𝑃(𝑋 ≥ 16) = 1 − 0.2375 = 0.7625 Step 2 e) The mean of binomial distribution is the product of the sample size n and the probability p: 𝜇 = 𝑛𝑝 = 20 × 0.20 = 14 f) The standard deviation of a binomial distribution is the square root of the product of the sample size n and the probabilities p and q. The variance is the square of the standard deviation. 𝜎 2 = 𝑛𝑝𝑞 = 𝑛𝑝(1 − 𝑝) = 20(0.70)(1 − 0.70) = 4.2 𝜎 = √𝑛𝑝𝑞 = √𝑛𝑝(1 − 𝑝) = √20(0.70)(1 − 0.70) ≈ 2.0494 Question #3 Simplify, please: 𝑛! (𝑛−3)! Answer: 𝑛! (𝑛 − 3)! 𝑛(𝑛 − 1)(𝑛 − 2)(𝑛 − 3)(𝑛 − 4) … .× 3 × 2 × 1 = (𝑛 − 3)(𝑛 − 4) … × 3 × 2 × 1 = 𝑛(𝑛 − 1)(𝑛 − 2) 𝑛(𝑛2 − 3𝑛 + 2) = 𝑛2 − 3𝑛2 + 2𝑛 Question #4 If a baseball player has a batting average of 0.275, what is the probability that the player will get the following number of hits in the next four times at bat? (A) Exactly 2 hits (B) At least 2 hits (A) P(exactly 2 hits) ≈ 0.029 (Round to three decimal places as needed.) Answer: Step 1 Introduction: Denote X as the random variable representing the number of hits made by the player in the next 4 times at bat. Step 2 A. Consider that the player hits or misses each of the next 4 times at bat, independently of each other. Further, consider it to be a success if the player makes a hit. As the batting average is 0.275, the probability of “success” is 𝑝 = 0.275 . Here, X can be considered as the number of successes observed among the 4 times at bat. Thus, it may be assumed that, X has a binomial distribution with parameters, 𝑛 = 4, 𝑝 = 0.275 . Since p denotes the probability of success, the probability of failure is, 𝑞 = 1 − 𝑝 = 0.725 . If 𝑋 ∼ 𝐵𝑖𝑛𝑜𝑚𝑖𝑎𝑙(𝑛, 𝑝), then the probability mass function of X is:  n x n− x ( ) p q ; x = 0,1,..., n; 0  p  1; q = 1 − p f ( x) =  x 0; otherwise  Here, n is the number of independent trials, p is the probability of success in each trial, and q is the probability of failure. The probability mass function of X here is: 4 x 4− x ( )(0.275) (0.725) ; x = 0,1, 2,3, 4. f ( x) =  x 0; otherwise  The probability that the player will get exactly 2 hits, that is, 𝑃(𝑋 = 2) is calculated below: 𝑃(𝑋 = 2) = 𝑓(2) 4 = ( )(0.275)2 (0.725) 4−2 2 = (6) ⋅ (0.075625) ⋅ (0.525625) ≈ 0.239 . Thus, the probability that the player will get exactly 2 hits is 0.239. Step 3 B. The probability that the player will get at least 2 hits, that is, 𝑃(𝑋 ≥ 2) is calculated below: 𝑃(𝑋 ≥ 2) = 𝑓(2) + 𝑓(3) + 𝑓(4) 4 4 4 = ( )(0.275)2 (0.725)4−2 + ( )(0.275)3 (0.725)4−3 + ( )(0.275)4 (0.725)4−4 2 3 4 ≈ (6) ⋅ (0.075626) ⋅ (0.525625) + (4) ⋅ (0.020797) ⋅ (0.725) + (1) ⋅ (0.2754) ⋅ (1) ≈ 0.238502 + 0.060311 + 0.005719 ≈ 0.305 . Thus, the probability that the player will get at least 2 hits is 0.305. Question #5 42% of adults say cashews are their favorite kind of nut. You randomly select 12 adults and ask each to name his or her favorite nut. Find the probability that the number who say cashews are their favorite nut is (a) exactlythree, (b) at least four, and (c) at most two. If convenient, use technology to find the probabilities. (a) 𝑃(3) = (b) 𝑃(𝑥 ≥ 4) = (c) 𝑃(𝑥 ≤ 2) = Answer: Step 1 Given Data: The probability that adults say that cashews are their favorite kind of nut is: 𝑝 = 0.43 The total number of adults randomly selected is: 𝑛 = 12 . The probability that the cashew nuts are not their favorite kind of nut is, 𝑞 =1−𝑝 Substitute values in the above expression. 𝑞 = 1 − 0.43 = 0.57 The binomial expression to calculate the probability of 'x' adults who says that cashews are their favorite kind of nut is, 𝑃(𝑥) =𝑛 𝐶𝑥 𝑝 𝑥 𝑞 𝑛−𝑥 Substitute values in the above expression. 𝑃(𝑥) =12 𝐶𝑥 × (0.43) 𝑥 × (0.57)12−𝑥 Step 2 (a) The expression to calculate the probability that exactly three adults say that the cashews are their favorite kind of nut is, 𝑃(3) =12 𝐶3 × (0.43)3 × (0.57)12−3 = 220 × (0.43)3 × (0.57)9 = 0.1111 Thus, the probability that exactly three adults say that the cashews are their favorite kind of nut is 0.111. (b) The expression to calculate the probability that at least four of the adults like cashews nuts. 𝑃(𝑥 ≥ 4) = 1 − (𝑃(0) + 𝑃(1) + 𝑃(2) + 𝑃(3)) Substitute values in the above expression. 𝑃(𝑥 ≥ 4) = 1 − (12 𝐶0 × (0.43)0 × (0.57)12 +12 𝐶1 × 0.43 × (0.57)11 +12 𝐶2 × (0.43)2 × (0.43)2 × (0.57)10 +12 𝐶3 × (0.43)3 × (0.57)9 = 1 − (0.0017 + 0.0106 + 0.0441 + 0.1110) = 1 − 0.1674 = 0.8326 Thus, the probability that at least four adults like cashews is0.8326. Step 3 (c) The expression to calculate the probability that at most two adults' like cashews is, 𝑃(𝑥 ≤ 2) = 𝑃(0) + 𝑃(1) + 𝑃(2) Substitute values in the above expression. 𝑃(𝑥 ≤ 2) =12 𝐶0 × (0.43)0 × (0.57)12 +12 𝐶1 × (0.43) × (0.57)11 +12 𝐶2 × (0.43)2 × (0.57)10 = 0.0017 + 0.0106 + 0.0441 = 0.0564 Thus, the probability that at most two adults like cashews is 0.0564. Question #6 An insurance salesman sells policies to 5 men the prob. that a mean will be a live in 30 years is 2 3 . Find the prob. that in 30 years: a) all 5 men , b) at least 3 men , c) only 2 men Answer: Let X be the number of men will be alive and n be sample number of men. From the given information, probability that a men will be alive in 30 years is 2 3 and 𝑛 = 5 . Here, men are independent and probability of success is constant. Hence, X follows binomial distribution 2 with parameters 𝑛 = 5 and 𝑝 = . 3 The probability mass function of binomial random variable X is n P( X = x) = ( ) p x (1 − p)n− x ; x = 0,1,..., n x Step 3 a. The probability that in 30 years all 5 men will be alive is 5 2 2 P( X = 5) = ( )( )5 (1 − )5−5 5 3 3 = 0.1317 Thus, the probability that in 30 years all 5 men will be alive is 0.1317. Step 4 b. The probability that in 30 years at least 3 men will be alive is 𝑃(𝑋 ≥ 3) = 𝑃(𝑋 = 3) + 𝑃(𝑋 = 4) + 𝑃(𝑋 = 5) 5 2 5 2 5 2 2 1 2 = ( )( )3 (1 − )5−3 + ( )( )4 (1 − )5−4 + ( )( )5 (1 − )5−5 3 3 4 3 5 3 3 2 3 = 0.3292 + 0.3292 + 0.1317 = 0.7901 The probability that in 30 years at least 3 men will be alive is 0.7901. Step 5 c. The probability that in 30 years all only 2 men will be alive is 5 2 2 P( X = 2) = ( )( )2 (1 − )5−2 2 3 3 = 0.1646 Thus, the probability that in 30 years all only 2 men will be alive is 0.1646. Question #7 Customers at a gas station pay with a credit card (A), debit card (B), or cash (C). Assume that successive customers make independent choices with 𝑃(𝐴) = 0.4, 𝑃(𝐵) = 0.2, and 𝑃(𝐶) = 0.4 . (a) Among the next 100 customers, what are the mean and variance of the number who pay with a debit card? mean customers variance customers 2 Explain your reasoning. Because we are interested in whether or not a debit card was used, we can use the binomial distribution. X = the probability that a customer used a debit card.Because we are interested in whether or not a debit card was used, we can use the binomial distribution. X = the number of customers who use a debit card. Because we are interested in whether or not a debit card was used, we can use the binomial distribution. X = the number of customers at the gas station. (b) Answer part (a) for the number among the 100 who don't pay with cash. mean customers variance customers 2 Answer: Step 1 Given that, Customers at a gas station pay with a credit card (A), debit card (B), or cash (C). Assume that successive customers make independent choices with 𝑃(𝐴) = 0.4, 𝑃(𝐵) = 0.2, and 𝑃(𝐶) = 0.4 . Reasoning: Because we are interested in whether or not a debit card was used, we can use the binomial distribution. Let X = the number of customers who use a debit card. Here, probability of customers who pay with debit card (or success) is 𝑝 = 0.2 Step 2 (a) Mean and variance of the number of customers who pay with a debit card: The mean number of customers who pay with a debit card can be calculated as given below: 𝜇 = 𝑛𝑝 = 100 × 0.2 = 20 The variance of the number of customers who pay with a debit card can be calculated as given below: 𝜎 2 = 𝑛𝑝(1 − 𝑝) = 100 × 0.2(1 − 0.2) = 16 Step 3 (b) Mean and variance of the number of customers who don't pay with cash: Here, probability of customers who don't pay with cash is 𝑝 = 0.6(= 1 − 0.4) The mean number of customers who don't pay with cash can be calculated as given below: 𝜇 = 𝑛𝑝 = 100 × 0.6 = 60 The variance of the number of customers who don't pay with cash can be calculated as given below: 𝜎 2 = 𝑛𝑝(1 − 𝑝) = 100 × 0.6(1 − 0.6) = 24 Question #8 Consider a binomial experiment with 𝑛 = 20 and 𝑝 = .70 . 1. Compute f (12). 2. Compute 𝑓(16). 3. Compute 𝑃(𝑥 ≥ 16) . Answer: Step 1 Since you have posted a question with multiple sub-parts, we will solve first three sub-parts for you. To get remaining sub-part solved please repost the complete question and mention the sub-parts to be solved.” (1) Compute f (12). The value of the function f (12) is obtained below: Let X denotes the random variable which follows binomial distribution with the probability of success 0.70 with the number trails 20. That is, 𝑛 = 12, 𝑝 = 0.70, 𝑞 = 0.30(= 1 − 0.70) The probability distribution is given by, n P( X = x) = ( ) p x (1 − p)n− x ; here x = 0,1, 2,..., n for 0  p  1 x Where n is the number of trials and p is the probability of success for each trial. The required probability is, 𝑓(12) = 𝑃(𝑋 ≤ 12) Use Excel to obtain the probability value for x equals 12. Follow the instruction to obtain the P-value: 1. Open EXCEL 2. Go to Formula bar. 3. In formula bar enter the function as“=BINOMDIST” 4. Enter the number of success as 12. 5. Enter the Trails as 20 6. Enter the probability as 0.70 7. Enter the cumulative as True. 8. Click enter EXCEL output: From the Excel output, the P-value is 0.2277 Thus, the value of the function f (12) is 0.2277. Step 2 2. Compute f (16). The value of the function f (16) is obtained below: The required probability is, f(16)=P(X \le 16) Use Excel to obtain the probability value for x equals 16. Follow the instruction to obtain the P-value: 1. Open EXCEL 2. Go to Formula bar. 3. In formula bar enter the function as“=BINOMDIST” 4. Enter the number of success as 16. 5. Enter the Trails as 20 6. Enter the probability as 0.70 7. Enter the cumulative as True. 8. Click enter EXCEL output: From the Excel output, the P-value is 0.8929 Thus, the value of the function f (16) is 0.8929. Step 3 (3) Compute the probability 𝑃(𝑥 ≥ 16) . The probability 𝑃(𝑥 ≥ 16) is obtained below as follows: The required probability is, 𝑃(𝑥 ≥ 16) = 1 − 𝑃(𝑥 < 16) = 1 − 𝑃(𝑥 ≤ 15) Use Excel to obtain the probability value for x equals 15. Follow the instruction to obtain the P-value: 1.Open EXCEL 2. Go to Formula bar. 3. In formula bar enter the function as“=BINOMDIST” 4. Enter the number of success as 15. 5. Enter the Trails as 20 6. Enter the probability as 0.70 7. Enter the cumulative as True. 8. Click enter EXCEL output: From the Excel output, the P-value is 0.7625. 𝑃(𝑥 ≥ 16) = 1 − 𝑃(𝑥 < 16) = 1 − 𝑃(𝑥 ≤ 15) = 1 − 0.7625 = 0.2375 The probability 𝑃(𝑥 ≥ 16) is 0.2375 . Question #9 Suppose we have two weighted coins, one of which comes up heads with probability 0.3, and the other of which comes up heads with probability 0.8. Unfortunately, the coins are otherwise identical, and we have lost track of which is which. Suppose we flip a randomly chosen coin 14 times and let N be the random variable giving the number of heads seen. If in the first 3 flips we see 2 heads, what is the conditional expected number of heads in the 14 flips? 𝐸[𝑁|2 of first 3 are H] =? Answer: Step 1 According to given data with the use of binomial distribution: 𝑃(𝑥) =𝑛 𝐶𝑥 𝑝 𝑥 (1 − 𝑝)𝑛−𝑥 Thus, Probability of coin 1 is: 𝑃(coin 1|2 heads in 3 flips) = 𝑃(coin 1|2 heads in 3 flips) = 𝑃(coin 1|2 heads in 3 flips) = P(coin 1 and 2 heads in 3 flips) P(2 heads in 3 flips) 3𝐶2 (0.3)2 (0.7) 3𝐶2 (0.8)2 (0.2)+3 𝐶2 (0.3)2 (0.7) 0.189 0.573 𝑃(coin 1|2 heads in 3 flips) = 0.3298 Step 2 The probability of coin 2 is: 𝑃(coin 2 and 2 heads in 3 flips) 𝑃(coin 2|2 heads in 3 flips) = } 𝑃(2 heads in 3 flips) 2 3𝐶2 (0.8) (0.2) 𝑃(coin 2|2 heads in 3 flips) = 2 3𝐶2 (0.8) (0.2) + 3𝐶2 (0.3)2 (0.7) 0.384 𝑃(coin 2|2 heads in 3 flips) = 0.384 + 0.189 0.384 𝑃(coin 2|2 heads in 3 flips) = 0.573 𝑃(coin 2|2 heads in 3 flips) = 0.6701 Now the probability of head is: 𝑃(𝐻𝑒𝑎𝑑𝑠) = (0.3 × 0.3298) + (0.8 × 0.6701) 𝑃(𝐻𝑒𝑎𝑑𝑠) = 0.0989 + 0.5360 𝑃(𝐻𝑒𝑎𝑑𝑠) = 0.6349 So 𝐸(𝑌) = 11 × 0.6349 = 6.98 𝐸(𝑋) = 2 + 6.98 = 8.98 Question #10 A binomial probability is given. Write the probability in words. Then, use a continuity correction to convert the binomial probability to a normal distribution probability: 𝑃(𝑥 > 73) Write the probability in words. The probability of getting _ 73 successes. Which of the following is the normal probability statement that corresponds to the binomial probability statement? A. 𝑃(72.5 < 𝑥 < 73.5) B. 𝑃(𝑥 < 72.5) C. 𝑃(𝑥 > 72.5) D. 𝑃(𝑥 < 73.5) E. 𝑃(𝑥 > 73.5) Answer: Step 1 Given: binomial probability is given by ; 𝑃(𝑥 > 73) Binomial probability in words : Probability of getting number of 73 successes . Step 2 Here we have to use continuity correction to convert binomial probability to normal probability as below; we know that; if we want to convert discrete to continuous then ; 𝑃(𝑥 > 𝑐) = 𝑃(𝑥 > 𝑐 + 0.5) therefore ; 𝑃(𝑥 > 73) = 𝑃(𝑥 > 73.5) ........ by using continuity correction Answer : Option 'E' is correct . Question #11 A binomial probability is given. Write the probability in words. Then, use a continuity correction to convert the binomial probability to a normal distribution probability: 𝑃(𝑥 > 116) Write the probability in words. The probability of getting _ 116 successes. Which of the following is the normal probability statement that corresponds to the binomial probability statement? A. 𝑃(𝑥 > 116.5) B. 𝑃(𝑥 > 115.5) C. 𝑃(115.5 < 𝑥 < 116.5) D. 𝑃(𝑥 < 115.5) E. 𝑃(𝑥 < 116.5) Answer: Step 1 Solution: Given information: The binomial probability is 𝑃(𝑥 ≤ 116) Step 2 The binomial probability in words is The probability of getting at most 116 successes . The continuity correction factor can be used to convert discrete probability into the continuous probability. By Central limit theorem the sampling distribution of sample mean approaches to the normal distribution if the sample size n gets larger (𝑛 > 30). In binomial distribution if both np and nq both greater than or equal to 10 or 5 then the binomial distribution can be approximated to the normal distribution. When we used continuity correction factor to convert discrete probability into continuous probability it is just adding or subtracting 0.5 to the discrete x value The given binomial probability can be converted to normal probability is 𝑃(𝑥 ≤ 116) use 𝑃(𝑥 < 116 + 0.5) 𝑃(𝑥 ≤ 116) = 𝑃(𝑥 < 116.5) The normal probability statement 𝑃(𝑥 < 116.5) is correspond to the binomial probability statement 𝑃(𝑥 ≤ 116) Option E is correct. Question #12 Assume that a procedure yields a binomial distribution with a trial repeated 𝑛 = 5𝑛 = 5 times. Use some form of technology to find the probability distribution given the probability 𝑝 = 0.6𝑝 = 0.6 of success on a single trial. (Report answers accurate to 2 decimal places.) k P( X = k ) 0 1 2 3 4 5 Answer: Step 1 From the given information, binomial distribution with a trial n=5 and the probability of success is p=0.6 . By using Excel formula to find the 𝑃(𝑋 = 0) For 𝑃(𝑋 = 0) : The excel formula for binomial probability is = BINOMDIST (Number_s, trials, probability_s,) The binomial probability is = BINOMDIST (0,5,0.6) Thus, the 𝑃(𝑋 = 0) is 0.01. For 𝑃(𝑋 = 1): The excel formula for binomial probability is = BINOMDIST (Number_s, trials, probability_s,) The binomial probability is = BINOMDIST (1,5,0.6) Thus, the 𝑃(𝑋 = 1) 𝑖𝑠 0.08. 𝑁 Step 2 For 𝑃(𝑋 = 2) : The excel formula for binomial probability is = BINOMDIST (Number_s, trials, probability_s,) The binomial probability is = BINOMDIST (2,5,0.6) Thus, the 𝑃(𝑋 = 2) is 0.23. For 𝑃(𝑋 = 3): The excel formula for binomial probability is = BINOMDIST (Number_s, trials, probability_s,) The binomial probability is = BINOMDIST (3,5,0.6) Thus, the 𝑃(𝑋 = 3) is 0.35. Step 3 For 𝑃(𝑋 = 4): The excel formula for binomial probability is = BINOMDIST (Number_s, trials, probability_s,) The binomial probability is = BINOMDIST (4,5,0.6) Thus, the 𝑃(𝑋 = 4) is 0.26. For 𝑃(𝑋 = 5) : The excel formula for binomial probability is = BINOMDIST (Number_s, trials, probability_s,) The binomial probability is = BINOMDIST (5,5,0.6) Thus, the 𝑃(𝑋 = 5) is 0.08. Step 4 The required probability values are tabulated below: k P( X = k ) 0 0.01 1 0.08 2 0.23 3 0.35 4 0.26 5 0.08 Question #13 X denotes a binomial random variable with parameters n and p. For each exercise, indicate which area under the appropriate normal curve would be determined to approximate the specified binomial probability. 𝑃(7 < 𝑋 < 10) Answer: Step 1 Continuity correction: The binomial probability is converted to a normal distribution probability by using the continuity correction. Step 2 If the binomial probability represents “ 𝑎 < 𝑋 < 𝑏 ” then add 0.5 to a and subtract 0.5 with b. That is, 𝑃(7 < 𝑋 < 10) = 𝑃(7 + 0.5 < 𝑥 < 10 − 0.5) = 𝑃(7.5 < 𝑥 < 9.5) Step 3 Thus, the area between 7.5 and 9.5 under the appropriate normal curve would estimate the binomial probability 𝑃(7 < 𝑋 < 10) . Question #14 In Exercises, X denotes a binomial random variable with parameters n and p. For each exercise, indicate which area under the appropriate normal curve would be determined to approximate the specified binomial probability. 𝑃(4 < 𝑋 < 8) Answer: Step 1 Continuity correction: The binomial probability is converted to a normal distribution probability by using the continuity correction. Step 2 If the binomial probability represents “ 𝑎 < 𝑋 < 𝑏 ” then add 0.5 to a and subtract 0.5 with b. That is, 𝑃(4 < 𝑋 < 8) = 𝑃(4 + 0.5 < 𝑥 < 8 − 0.5) = 𝑃(4.5 < 𝑥 < 7.5) Step 3 Thus, the area between 4.5 and 7.5 under the appropriate normal curve would estimate the binomial probability 𝑃(4 < 𝑋 < 8) . Question #15 X denotes a binomial random variable with parameters n and p, Indicate which area under the appropriate normal curve would be determined to approximate the specified binomial probability. 𝑃(𝑋 ≥ 8) Answer: Step 1 Continuity correction: The binomial probability is converted to a normal distribution probability by using the continuity correction. If the binomial probability represents “ 𝑋 ≥ 𝑎 ” then subtract 0.5 from a and 0.5 to n where n represents the number of trials. Step 2 That is, 𝑃(𝑋 ≥ 8) = 𝑃(8 − 0.5 < 𝑥 < 𝑛 + 0.5) = 𝑃(7.5 < 𝑥 < 𝑛 + 0.5) Thus, the area between 7.5 and 𝑛 + 0.5) under the appropriate normal curve would estimate the binomial probability 𝑃(𝑋 ≥ 8) . Question #16 A binomial probability distribution has 𝑝 = .20 and 𝑛 = 100 . a. what are the mean and standard deviation? b. Is this situation one in which binomial probabilities can be approximated by the normal probability distribution? Explain. c. what is the probability of exactly 24 successes? d. what is the probability of 18 to 22 successes? e. what is the probability of 15 or fewer successes? Answer: Step 1 a. The binomial probability distribution has 𝑝 = 0.20 and 𝑛 = 100 . Assume that x is a binomial random variable with 𝑝 = 0.20 and 𝑛 = 100 . The formula for mean in binomial distribution is, 𝜇 = 𝑛𝑝 , Where p is the probability of success and n is the number of trials. The formula for standard deviation is, ∑ = √𝑛𝑝(1 − 𝑝). The value of mean is, 𝜇 = 100 × 0.20 = 20 Thus, the mean is 20. The value of standard deviation is, 𝜎 = √100 × (0.20 × 0.80) = √16 = 4 Thus, the standard deviation is 4. Step 2 b. The requirements to check whether the normal distribution can be used to approximate the binomial distribution are: Condition 1: 𝑛𝑝 ≥ 5 Condition 2: 𝑛(1 − 𝑝) ≥ 5 Condition 1: 𝑛𝑝 ≥ 5 Substitute 100 for n and 0.20 for p in the np. 𝑛𝑝 = 100 × 0.20 = 20 > 5 Thus, the requirement 𝑛𝑝(= 20) ≥ 5 is satisfied. Condition 2: 𝑛(1 − 𝑝) ≥ 5 Substitute 100 for n and 0.20 for p in the 𝑛(1 − 𝑝). 𝑛(1 − 𝑝) = 100 × (1 − 0.20) = 100 × 0.80 = 80 > 5 Thus, the requirement 𝑛𝑞(= 80) ≥ 5 is satisfied. Thus, the normal distribution can be used to approximate the binomial distribution. Step 3 c. Continuity correction: The binomial probability is converted to a normal distribution probability by using the continuity correction. If the binomial probability represents “exactly c” then the normal probability is P(c-0.5