Approximately 70% of U.S. adults had at least one pet as a child. We randomly survey 60 U. S. adults. We are interested in the n

Question

3 years ago

11

Approximately 70% of U.S. adults had at least one pet as a child. We randomly survey 60 U. S. adults. We are interested in the n

umber that had at least one pet as a child. The probability that at least 3 adults had at least one pet as a child means:
A. P(X=0)+P(X=1)+P(X=2)+P(X=3)
B. P(X=0)+P(X=1)+P(X=2)
C. P(X=4)+P(X=5)+P(X=6)+ ...
D. P(X=3)+P(X=4)+P(X=5)+ ...

Mathematics

2 answers:

Kryger [21] · Answer 1 · 2022-11-07T21:36:59+03:00

Kryger [21]3 years ago

6 0

Answer:

D. P(X=3)+P(X=4)+P(X=5)+ ...

ipn [44] · Answer 2 · 2022-11-07T20:34:10+03:00

ipn [44]3 years ago

5 0

Answer:

D. P(X=3)+P(X=4)+P(X=5)+ ...

Step-by-step explanation:

Given

$n =60$

$pr = 70\%$ -- proportion of adults with pet

Required

Represent at least 3 adult with pet as a probability

At least 3 means 3 or more than.

So, the probability is represented as:

$P(x \ge 3) = P(3) + P(4) + P(5) + ........$

Hence;

<em>(d) is correct</em>