The following data represent the ages of the best across for the years 1985-2004

Question

The following data represent the ages of the best across for the years 1985-2004. 39 49 62 52 39 47 38 32 42 51 25 30 53 44 51 48 61 35 37 42 Give the four measures of central tendency. A standard deck of cards has 52 cards. One card is selected at random from the deck. Compute. Probability of randomly selecting a nine or eight or three. Probability of randomly selecting a ten or club According to almanac, 60% of adult smokers stated smoking before turning 18 years old. Compute the mean and the standard deviation of the random variable X, the number of smokers who started before 18 m 300 trials. Would it be unusual to observe 255 smokers who started before turning 18 years old in random sample of 300 adult smokers? why?

Explanation / Answer

Given that the standard deck has 52 cards.

One card is selected at random from the deck.

Here we have to find P(selecting a 9 or 8 or 3).

There are 4 cards of 9, 4 cards of 8 and 4 cards of 3.

By using addition rule of probability,

P(selecting a 9 or 8 or 3) = P(selecting a 9) + P(selecting a 8) + P(selecting a 3) - Intersection probability.

But intersection probability is 0 because the events are independent.

P(selecting a 9 or 8 or 3) = 4/52 + 4/52 + 4/52 - 0 = 12/52 with replacement

And for without replacement,

P(selecting a 9 or 8 or 3) = 4/52 + 4/51 + 4/50 = 0.2354

P(selecting a 10 or club) = P(selecting 10) + P(club) - P(selecting 10 and club)

There are 4 cards of 10.

There are 13 cards of club.

And 1 card of 10 and club.

P(selecting a 10 or club) = 4/52 + 13/51 - 1/50 = 0.3118

Given that,

p = proportion of adult smokers started smoking before turning 18 years old = 60% = 0.60

n = number of trials = 300

We go from binomial to normal if np > 10 and nq > 10.

q = 1-p

q = 1-p = 1-0.6 = 0.4

np = 300*0.6 = 180

nq = 300*0.4 = 120

Both np and nq > 10.

sample proportion p^ ~ N(mean = p, sd = sqrt((pq)/n) )

mean = 0.6

sd = sqrt((0.6*0.4)/300) = 0.0283

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