Among 46- to 51-year-olds, 28% say they have called a talk show while under the influence of peer pressurepeer pressure. Supp
1 answer:
Answer:
0.8997, 0.9999
Step-by-step explanation:
Given that among 46- to 51-year-olds, 28% say they have called a talk show while under the influence of peer pressure.
i.e. X no of people who say they have called a talk show while under the influence of peer pressure is binomial with p = 0.28
Each person is independent of the other and there are only two outcomes
n =7
a) the probability that at least one has not called a talk showcalled a talk show while under the influence of peer pressure
= where Y is binomial with p = 0.72
=
b) the probability that at least one has called a talk showcalled a talk show while under the influence of peer pressurepeer pressurethe probability that at least one has not called a talk showcalled a talk show while under the influence of peer pressure
=
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Answer:
a) 37.31 b) 42.70 c) 0.57 d) 0.09
Step-by-step explaanation:
We are regarding a normal distribution with a mean of 35 and a standard deviation of 6, i.e., = 35 and
= 6. We know that the probability density function for a normal distribution with a mean of
and a standard deviation of
is given by
in this case we have
Let be the random variable that represents a row score, we find the values we are seeking in the following way
a) we are looking for a number such that
=
, this number is
=37.31
you can find this answer using the R statistical programming languange and the instruction qnorm(0.65, mean = 35, sd = 6)
b) we are looking for a number such that
=
, this number is
=42.70
you can find this answer using the R statistical programming languange and the instruction qnorm(0.9, mean = 35, sd = 6)
c) we find this probability as
=
you can find this answer using the R statistical programming languange and the instruction pnorm(38, mean = 35, sd = 6) -pnorm(28, mean = 35, sd = 6)
d) we find this probability as
=
you can find this answer using the R statistical programming languange and the instruction pnorm(44, mean = 35, sd = 6) -pnorm(41, mean = 35, sd = 6)