Can som1 plz help me solve this statistics problems? I'm stocked on these problems.
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Can som1 plz help me solve this statistics problems? I'm stocked on these problems.

[From: ] [author: ] [Date: 11-11-28] [Hit: ]
the final exam is 25% of the final grade, and the homework grade is 15% of the final grade. What is the students mean score in the class?Plz,p = probability that an individual is left-handed = 0.q = probability that an individual is not left-handed = 1 - 0.......
1) A multiple-choice test has five questions, each with five choices for the answer. Only one of the choices is correct. You randomly guess the answer to each question. What is the probability that you answer the first two questions correctly?

2)The probability that an individual is left-handed is 0.17. In a class of 40 students, what is the mean and standard deviation of the number of left-handers in the class?

3)A card is drawn from a standard deck of 52 playing cards. Find the probability that the card is an ace or a black card.

4)A recent survey found that 70% of all adults over 50 wear glasses for driving. In a random sample of 10 adults over 50, what is the probability that at least six wear glasses?

5)A student receives test scores of 62, 83, and 91. The student's final exam score is 88 and homework score is 76. Each test is worth 20% of the final grade, the final exam is 25% of the final grade, and the homework grade is 15% of the final grade. What is the student's mean score in the class?

Plz, show the working
Thnks in advance

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1)

Probability of answering any question correctly = 1/5

Probability of answering first 2 questions correctly = 1/5 * 1/5 = 1/25

--------------------

2)

p = probability that an individual is left-handed = 0.17
q = probability that an individual is not left-handed = 1 - 0.17 = 0.83
n = number of students in class = 40

mean = n * p = 40 * 0.17 = 6.8
variance = n * p * q = 40 * 0.17 * 0.83 = 5.644
standard deviation = √5.644 = 2.3757

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3)

P(A or B) = P(A) + P(B) - P(A and B)

P(card is an ace or a black card)
= P(card is an ace) + P(card is a black card) - P(card is a black ace)
= 4/52 + 26/52 - 2/52
= 28/52
= 7/13

Alternatively, we know there are 26 black cards in deck
Of the red cards, 2 are aces
(26+2)/52 = 28/52 = 7/13

--------------------

4)

This is a binomial distribution with p = 7/10, q = 3/10

P(X = x) = Probability that x of the 50 randomly chosen wear glasses
P(X = x) = (50Cx) p^x q^(50-x) = (50Cx) (7/10)^x (3/10)^(50-x)

P(at least 6 wear glasses)
= P(X ≥ 6)
= 1 - P(X < 6)
= 1 - [P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)]
= 0.999999999999997

--------------------

5)

Mean score = 0.2(62) + 0.2(83) + 0.2(91) + 0.25(88) + 0.15(76)
. . . . . . . . . . = 80.6

Mαthmφm
1
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