Which situation involves descriptive statistics?

Mathematics

2 answers:

irinina [24]2 years ago

7 0

Answer:

Step-by-step explanation:

Descriptive statistics describe an event that happens over time. So bowler striking on 1/5 of his throws that night would be an example of descriptive statistics.

Ghella [55]2 years ago

7 0

(c)A bowler’s scorecard shows he threw a strike on one-fifth of his throws that night. Descriptive statistics describe an event that happens over time, for example, a batting average would be a descriptive statistic or a win/loss ratio.

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Thirty percent of all customers who enter a department store will make a purchase. Suppose that 6 customers enter the store and

torisob [31]

Answer:

Step-by-step explanation:

Let's use binomial distribution, because we are interested in find the total number of successes in a sequence of n independent experiments.

In probability, a binomial distribution is used in a Bernoulli process. That is, a sequence of experiments of n independent trials, each asking a dichotomous question, that means it only has two answers. One answer is a success (probability: p) and the other is a failure (probability: 1-p). In this case, every customer who enters in the store is the experiment. If they make a purchase is a success event, taking into account that every purchase decision has to be independent.

$P(X=x) = C(n,x)p^{x} (1-p)^{n-x}$

x= total number of success

n= total number of experiments

p=probability of success on an indivual trial

$C(n,x) = \frac{n!}{x!(n-x)!}$

a) x= 5 ------> customers that make a purchase

n= 6 ------> total customers

p=0.3

$p(X=5)= C(6,5)(0.3)^{5} (1-0.3)^{6-5} \\p(X=5)=0.102$

b) At least 3 customers make a purchase. x ≥ 3

$P(X\geq 3)=1-P (X=0)-P(X=1)-P(X=2)\\P(X\geq 3)=1-(C(6,0)(0.3^{0})(0.7)^{6}) - (C(6,1)(0.3^{1})(0.7)^{5}) - (C(6,2)(0.3^{2})(0.7)^{4})\\ P(X\geq 3) = 1-0.11765-0.30253-0.32414\\P(X\geq3)= 0.2557$

c) At most 2 customers make a purchase. x ≤ 2

$P(X\leq2) = P(X=0) + P(X=1) + P(X=2)\\P(X\leq2) = (C(6,0)(0.3^{0})(0.7)^{6}) + (C(6,1)(0.3^{1})(0.7)^{5}) + (C(6,2)(0.3^{2})(0.7)^{4})\\P(X\leq2) = 0.11765-0.30253-0.32414\\P(X\leq2)=0.7443$