I need help with number 14 as soon as possible i have class in two hours please

Question

I need help with number 14 as soon as possible i have class in two hours please help.

h of three, he wins $20 ) $3.06 " this garhe? (2pts) D) $2.78 -83.06. s. How many different four-letter code words can be fomed from the l eaegoys (2pts) 9. Using the following, construct a complete frequency distributice include tde mean, Variusce. Standard Deviation. Also graph the Frequency Histognam, Polygon and Ogive. Indicste Modal Class, Median Class and describe the graph (20 points) Class Boundaries Frequency 28.5-33.5 33.3-38.3 38.5-43.5 43.5-48.5 48.5-53.3 10 10. Given twelve students, five of which are females, if two students are selected at random, without replacement, what is the probability that both students are female? (2pts) 11. Find the weighted mean for three exams if the irst one was worth 20% and the student received a the second was worth 50 % and the student received a 84 and the third was worth 30 % and the student received an 73? (2pts highway are traveling at speeds in excess of 70 mph. If It has been reported that 25% of all cars on the speeds of eight random autamobiles are measured vis radar: (3pts each 6 poims) 12. a What is the probability that exactly 3 cars are speeding? b. What is the probability that at least one car is poing over 70 mph 13. Using the probability distribution listed below (3yts each 12 points) P(x) 0.1s 0.20 0.30 025 0.10 a. Find the mean, variance and standard deviation b. Find Pix is no more than 3)? c. Find P(x is ad least 2)? d. Find Pix 2 orx-2)? 14. It has been reported that 60% of all intemet users have been anacked by viruses randomly surveyed, find the following probabilities; (pts cach- 12 pes) iri 6 people are a. At least 6 have been allacked b. No more than 4 have been attacked c. Less thin 13 have been attacked Page 2

Explanation / Answer

Question 14

Solution:

Here, we have to use normal approximation for binomial distribution.

We are given,

n = 16

p = 0.60

so, q = 1 – p = 1 – 0.60 = 0.40

n*p = 16*0.60 = 9.6

n*q = 16*0.40 = 6.4

So, n*p > 5 and n*q > 5

So, we can use normal approximation.

Mean and SD for normal approximation are given as below:

Mean = n*p = 16*0.60 = 9.6

SD = sqrt(n*p*q) = sqrt(16*0.60*0.40) = 1.959592

Part a

We have to find P(X6)

By subtracting continuity correction factor 0.5, we have to find P(X5.5)

P(X5.5) = 1 – P(X<5.5)

Z = (X – mean) / SD

Z = (5.5 – 9.6) / 1.959592

Z = -2.09227

P(Z<-2.09227) = P(X<5.5) = 0.018207

(By using z-table or excel)

P(X5.5) = 1 – P(X<5.5)

P(X5.5) = 1 – 0.018207

P(X5.5) = 0.981793

Required probability = 0.981793

Part b

Here, we have to find P(X4)

By adding continuity correction factor 0.5, we have to find P(X4.5)

Z = (4.5 - 9.6) / 1.959592

Z = -2.60258

P(Z<-2.60258) = P(X4.5) = 0.004626

(By using z-table or excel)

Required probability = 0.004626

Part c

We have to find P(X<13)

By subtracting continuity correction 0.5, we have to find P(X<12.5)

Z = (12.5 - 9.6) / 1.959592

Z = 1.4799

P(Z<1.4799) = P(X<12.5) = 0.93055

(By using z-table or excel)

Required probability = 0.93055

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