What do you mean by Spamming?

Spamming means is when someone annoy the person, by putting gibberish words or by trying to get someone attentions.

Hopefully, this helps you!!

Step-by-step explanation:

1)The probability of success in each of the 58 identical engine tests is p=0.92

n = 58

mean, u = np = 58×0.92 = 53.36

2) The only value that would be considered usual for this distribution is 91. This is because it is the only value between the minimum and maximum value

3) n = 546

p = 17/100 = 0.17

Mean = np = 546×0.17= 92.82

4) n = 1035

p = 36/100 = 0.36

np = 1035 × 0.36 = 372.6

5) The probability of success is 0.2.

p = 0.2

q= 1-p = 1-0.2 = 0.8

n = 35

standard deviation =

√npq = √35×0.2×0.8 = 2.34

6) p = 0.25

q = 1-0.25 = 0.75

n = 5

Variance = npq = 5×0.25×0.75 = 0.9

7) n = 982

p = 0.431

q = 1 - p = 1 - 0.431 = 0.569

Variance = npq = 982×0.431×0.569= 240.8

8) n = 500

p = 84/100 = 0.84

q = 1-0.84 = 0.16

Standard deviation = √npq

Standard deviation = √500×0.84×0.16 = 8.2

Step-by-step explanation:

1)The probability of success in each of the 58 identical engine tests is p=0.92

n = 58

mean, u = np = 58×0.92 = 53.36

2) The only value that would be considered usual for this distribution is 91. This is because it is the only value between the minimum and maximum value

3) n = 546

p = 17/100 = 0.17

Mean = np = 546×0.17= 92.82

4) n = 1035

p = 36/100 = 0.36

np = 1035 × 0.36 = 372.6

5) The probability of success is 0.2.

p = 0.2

q= 1-p = 1-0.2 = 0.8

n = 35

standard deviation =

√npq = √35×0.2×0.8 = 2.34

6) p = 0.25

q = 1-0.25 = 0.75

n = 5

Variance = npq = 5×0.25×0.75 = 0.9

7) n = 982

p = 0.431

q = 1 - p = 1 - 0.431 = 0.569

Variance = npq = 982×0.431×0.569= 240.8

8) n = 500

p = 84/100 = 0.84

q = 1-0.84 = 0.16

Standard deviation = √npq

Standard deviation = √500×0.84×0.16 = 8.2

In case anyone needs confirmation, I just took the quiz and it is indeed D,

"a word or expression that is new or has a new use".

Hope I could help, good luck :)

a) 0.941

b) 0.593

c) 0.02901

d) 0.9999999999

Step-by-step explanation:

A student has two accounts.

70% of her messages come from account 1.

Probability that a message is on account 1 = P(A) = 0.7

Meaning 30% of her messages come from account 2.

Probability that a message is on account 2 is P(B) = 0.3

Of the messages that come into Account 1, 5% are spam

Calculating in terms of total percentage of messages.

The amount of spam on account 1 is 5% of 70% of the total amount of messages.

P(A n S) = 0.05 × 0.7 = 0.035 = 3.5%

The amount of spam on account 2 is 8% of 70% of the total amount of messages.

P(B n S) = 0.08 × 0.3 = 0.024 = 2.4%

Total amount of spam messages received = P(S) = 2.4% + 3.5% = 5.9% = 0.059

a) If a message is randomly selected, what is the probability that it is NOT a spam?

P(S') = 1 - P(S) = 1 - 0.059 = 0.941

b) If a randomly selected message is a spam, what is the probability that it came from Account 1.

P(A|S) = P(A n S)/P(S)

P(A|S) = (0.035/0.059)

P(A|S) = 0.593

c) Suppose 5 messages are randomly selected. What is the probability that exactly 2 of them are spam

This is a binomial distribution problem

Binomial distribution function is represented by

P(X = x) = ⁿCₓ pˣ qⁿ⁻ˣ

n = total number of sample spaces = 5 messages picked

x = Number of successes required = number of spam messages that should be amongst the 5 messages picked = 2

p = probability of success = probability that a message is a spam message = 0.059

q = probability of failure = probability that a message is NOT a spam message = 0.941

P(X=2) = ⁵C₂ (0.059)² (0.941)⁵⁻²

P(X=2) = 0.02901

d) Suppose 20 messages are randomly selected. What is the probability that at least 3 of them are not spam?

This is also a binomial distribution problem

Binomial distribution function is represented by

P(X = x) = ⁿCₓ pˣ qⁿ⁻ˣ

n = total number of sample spaces = number of messages = 20

x = Number of successes required = at least 3

p = probability of success = probability that a message is not a spam = 0.941

q = probability of failure = probability that a message is a spam message = 0.059

P(X ≥ 3) = 1 - P(X < 3)

P(X < 3) = P(X=0) + P(X=1) + P(X=2) = 0.0000000000001

P(X ≥ 3) = 1 - P(X < 3) = 1 - 0.0000000000001 = 0.9999999999 (Almost 1!)

Hope this Helps!!!

a) 0.941

b) 0.593

c) 0.02901

d) 0.9999999999

Step-by-step explanation:

A student has two accounts.

70% of her messages come from account 1.

Probability that a message is on account 1 = P(A) = 0.7

Meaning 30% of her messages come from account 2.

Probability that a message is on account 2 is P(B) = 0.3

Of the messages that come into Account 1, 5% are spam

Calculating in terms of total percentage of messages.

The amount of spam on account 1 is 5% of 70% of the total amount of messages.

P(A n S) = 0.05 × 0.7 = 0.035 = 3.5%

The amount of spam on account 2 is 8% of 70% of the total amount of messages.

P(B n S) = 0.08 × 0.3 = 0.024 = 2.4%

Total amount of spam messages received = P(S) = 2.4% + 3.5% = 5.9% = 0.059

a) If a message is randomly selected, what is the probability that it is NOT a spam?

P(S') = 1 - P(S) = 1 - 0.059 = 0.941

b) If a randomly selected message is a spam, what is the probability that it came from Account 1.

P(A|S) = P(A n S)/P(S)

P(A|S) = (0.035/0.059)

P(A|S) = 0.593

c) Suppose 5 messages are randomly selected. What is the probability that exactly 2 of them are spam

This is a binomial distribution problem

Binomial distribution function is represented by

P(X = x) = ⁿCₓ pˣ qⁿ⁻ˣ

n = total number of sample spaces = 5 messages picked

x = Number of successes required = number of spam messages that should be amongst the 5 messages picked = 2

p = probability of success = probability that a message is a spam message = 0.059

q = probability of failure = probability that a message is NOT a spam message = 0.941

P(X=2) = ⁵C₂ (0.059)² (0.941)⁵⁻²

P(X=2) = 0.02901

d) Suppose 20 messages are randomly selected. What is the probability that at least 3 of them are not spam?

This is also a binomial distribution problem

Binomial distribution function is represented by

P(X = x) = ⁿCₓ pˣ qⁿ⁻ˣ

n = total number of sample spaces = number of messages = 20

x = Number of successes required = at least 3

p = probability of success = probability that a message is not a spam = 0.941

q = probability of failure = probability that a message is a spam message = 0.059

P(X ≥ 3) = 1 - P(X < 3)

P(X < 3) = P(X=0) + P(X=1) + P(X=2) = 0.0000000000001

P(X ≥ 3) = 1 - P(X < 3) = 1 - 0.0000000000001 = 0.9999999999 (Almost 1!)

Hope this Helps!!!