This problem will ask you to use appropriate functions of R to answer probabilit

Question

This problem will ask you to use appropriate functions of R to answer probability questions similar to the ones we have be doing in class. (Hnnnm...I wonder if these objects should be saved and named exactly as indicated??) (a) The amount of ketchup dispensed from a machine at a local convenience store is normally distributed with a mean of 1.10 ounces and a standard deviation of 0.085 ounce. Use the pnorm function to determine the probability that the machine dispenses more than 1 ounce for a giving serving? Assign that value to the object 4x 5a'. (b) The scores on the SAT are normally distributed. The mean SAT score during a certain year was 995 with a standard deviation of 175. Use the qnorm function to determine the cutoff score for the 88 percentile. Assign that value to the object 'x5b'. (c) Use the dbinom function to compute the probability of a (fair) coin landing heads exactly 92 times in 153 flips of the coin. Assign that value to the object 'x5c'. (d) A student Is taking a 30-question multiple-choice quiz, each with 5 answer choices (only one of which is correct). Use the pbinom function to determine the probability that the student answers no more than 9 questions correct if that student randomly guesses the answer to every question on the quiz. Assign that value to the object 'x5d' (e) A six-sided die Ls biased so that the probability of rolling a 4 on the single roll of the die Ls only 10%. Use the following while loop to determine the minimum number to times the die would need to be rolled so that the probability of rolling a total of fifty or more 4's is at least 90%. Assign that minimum number of rolls to the object 'x5e' > n=50 #We can start with n = 50 rolls since we are looking for at least 50 successes. > while(1-pbinoin(49. n, .10)

Explanation / Answer

a) x5a <- pnorm(1, mean=1.1, sd=0.085)
> x5a
[1] 0.1197034

b)x5b <- qnorm(0.88, mean=995, sd=175)
> x5b
[1] 1200.623

c)x5c <- dbinom(92, 153, 0.5)
> x5c
[1] 0.002783212

d) x5d <- pbinom(9,30,0.2)
> x5d
[1] 0.9389129

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