On the occasion of its 10-year anniversary, AJ Inc. sells lucky draw tickets. If the customers are interested in the lucky draw contest, then they have to purchase a ticket for $15. The gift could be worth $115, $215, $315, or nothing. The probability of each event is given below. What is the value of the standard deviation

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On the occasion of its 10-year anniversary, AJ Inc. sells lucky draw tickets. If the customers are interested in the lucky draw contest, then they have to purchase a ticket for $15. The gift could be worth $115, $215, $315, or nothing. The probability of each event is given below. What is the value of the standard deviation

Probability              0.35     0.26     0.21      0.18

Amount gained     $100     $200   $300    -$15

Option;

a) $11,143

b) $106

c) $10,200

d) $147

the standard deviation is 106

Option b) $106 is the correct Answer

Step-by-step explanation:

Given the data in the question;

let the random variable be x

x = amount gained by customers in a lucky draw contest

so

Variance (x) = [∑(x² × P(x))] - [(∑(x × P(x))²]

so

x              x²               p(x)        x.p(x)       x².p(x)

$100      10,000        0.35       35           3,500

$200     40,000       0.26       52          10,400

$300     90,000       0.21        63           18,900

-$15         225           0.18       -2.7          40.5

TOTAL                                   147.3         32,840.5

Variance (x) = [∑(x² × P(x))] - [(∑(x × P(x))²]

we substitute;

Variance (x) = 32,840.5 - (147.3)²

Variance (x) = 32,840.5 - 21,697.29

Variance (x) = 11,143.21

Now Standard Deviation = √variance

so, S.D = √11,143.21

S.D = 105.56 ≈ 106

Therefore, the standard deviation is 106

Option b) $106 is the correct Answer

i need points

step-by-step explanation:

sorry bubs