It is known that 10% of the calculators shipped from a particular factory are defective. What is the probability that no more than one in a random sample of four calculators is defective?

0.9477

Step-by-step explanation:

p = 10% = 0.1, q = 90% = 0.9, n = 4

The question follows a binomial probability distribution since the experiment (defectiveness of random sample of calculator) is performed more than once ( 4 calculators are defective).

The question is to find the probability that not more than one in a random sample of 4 calculators is defective that's

p(x≤1) = p(x=0) + p(x=1)

The probability mass function of a binomial probability distribution is given below as

P(x=r) =nCr × p^r × q^n-r

At x = 0

p(x=0) = 4C0 × 0.1^0 × 0.9^4-0

p(x=0) = 4C0 × 0.1^0 × 0.9^4

p(x=0) = 1 × 1 × 0.6561

p(x=0) = 0.6561.

At x = 1

p(x=1) = 4C1 × 0.1^1 × 0.9^4-1

p(x=1) = 4C1 × 0.1^1 × 0.9^3

p(x=1) = 4 × 0.1 × 0.729

p(x=1) = 0.2916

p(x≤1) = 0.6561 + 0.2916 = 0.9477

not enough information

step-by-step explanation:

we cannot know this, because the other students could have handed out 10 flyers, or they could have handed out 100. one way, the mean would go up, and another way, the mean would go down.

8

step-by-step explanation:

to find the amount of gift bags to use all the supplies, find the gcf or greatest common factor. this is the largest number which multiplies evenly into each term.

48 has factors: 1, 2, 3, 4, 6, 8, 12. 16, 24, 48

80 has factors: 1, 2, 4, 5, 8, 10, 16, 20, 40, 80

64 has factors: 1, 2, 4, 8, 8, 16, 32, 64

the highest number factor they all share in common is 8. this means there will be 8 bags with 6 pencils, 10 markers, and 8 stickers in each bag.