1. Use the appropriate table to solve this problem: 30% of hedgehog owners choos

rocedwrp

rocedwrp

Answered question

2021-09-20

1. Use the appropriate table to solve this problem: 30% of hedgehog owners choose fresh food to feed their pets (the rest chooses dried cat food).
If 8 hedgehog owners are chosen at random, what is the probability that at most 2 of them feed their pets fresh food?
Fill in the following list: n=p=
Then work the problem:
(give the answer to four decimal places)

Answer & Explanation

SchepperJ

SchepperJ

Skilled2021-09-21Added 96 answers

Step 1
Obtain the probability that at most 2 of them feed their pets fresh food.
The probability that at most 2 of them feed their pets fresh food is obtained below as follows:
Let X denotes the number of the hedgehog owners choose fresh food to feed their pet which follows binomial distribution with the probability of success 0.30 and the random sample of owners selected is 8.
That is, n=8,p=0.30.
The probability distribution is given by,
P(X=x)=(beg{array}{c}8xend{array})0.30x(10.30)8x; here x=0,1,2,,8 for 0p1
Where n is the number of trials and
p is the probability of success for each trial.
The required probability is,
P(X2)=P(X=0)+P(X=1)+P(X=2)
Step 2
For X=0,n=8,p=0.30
From the “Binomial probability distribution table”, at the values n=8,x=0 with the probability of success p=0.30 the probability value is 0.0576
For X=1,n=8,p=0.30
From the “ Binomial probability distribution table”, at the values n=8,x=1 with the probability of success p=0.30 the probability value is 0.1977
For X=2,n=8,p=0.30
From the “Binomial probability distribution table”, at the values n=8,x=2 with the probability of success p=0.30 the probability value is 0.2965.
Therefore,
P(X2)=P(X=0)+P(X=1)+P(X=2)
=0.0576+0.1977+0.2965
=0.5518
The probability that at most 2 of them feed their pets fresh food is 0.5518.

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