Suppose 60% of a herd of cattle is infected with a particular disease. Let X= th

Question

Suppose 60% of a herd of cattle is infected with a particular disease. Let X= the number of infected and Y= the number of non-infected cattle in a sample of size 5. The distribution of Y is binomial with n = 5 and p = 0.6 binomial with n = 5 and p = 0.4 binomial with n = 5 and p = 0.5 the same as the distribution of X. the number of infected If one card is drawn from a deck, find the probability of getting these result A club card 1/52 4/52 8/52 13/52 12/52 A 6 and a spade 1/52 4/52 8/52 13/52 12/52 A card less than 6 (an ace is considered above 6) 1/52 16/52 12/52 8/52 4/52 Select the best answer for each of the following: If a group of subjects have a mean score of 20 and a standard deviation of 4 on a test, approximately. 95% of the scores lie between: X is a normally distributed random variable with a mean of 22 and a standard deviation of 5. The probability that X is less than 9.7 is 0.0000 0.4931 00.0069 0.9931 0.0505 0.1550

Explanation / Answer

10.) There are 13 club cards in a deck of cards.

The probability for selecting one of the 13 cards from 52 cardsis, therefore, 13/52

Answer : d

11.) There is only one six card which has a spade. So the probability of selecting this one card from 52 cards is 1/52.

Answer : a

12.) The cards less than six are 2, 3 , 4 and 5

There are 4 cards of each of these.

Therefore the probability of selecting from 12 cards out of 52 is 12/52.

Answer : c

Q2.)

1.) 95% of the scores will lie between two standard deviations of the mean

i.e. (20-8 , 20+8)

12 and 28

Answer : d

2.) P( X < 9.7)

= P( Z < (9.7 - 22)/5)

= P( Z < -2.46)

= 0.0069

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