The amount of mustard dispensed from a standard dispenser is normally distributed with a mean of 0.9oz and a variance of 0.01oz. If a family of six buys ten hotdogs total and dispenses mustard onto each one using the standard dispenser, what is the probability that more than 3 of them will have more than one ounce of mustard?

the probability that more than 3 of them will have more than one ounce of mustard is 0.05975

Step-by-step explanation:

Given te data in the question;

let x rep the quantity of mustard on a hotdog ( in oz)

so

X → N ( u = 0.9, α² = 0.01 )

now P( more than one ounces of mustard on a hotdog) will be

P( X>1 ) = P( Z > ((1-0.9)/(√0.01)) ) = P( Z > 1 ) = 1 - P( Z < 1 )

= 1 - 0.84134 = 0.15866

Also let Y rep the number of hotdogs that have more than one ounces of mustard, so

Y → Bin( n = 10, p = 0.15866 )

P( y = y ) = (¹⁰ _y) ( 0.15855)⁰ ( 1 - 0.15866 )^10-y, y = 0,1,2,3, 10

so required probability = P(Y > 3) = 1 - P(Y ≤ 3)

= 1 - P(Y=0) - P(Y=1) - P(Y=2) - P(Y=3)

=1 -  (¹⁰ ₀) ( 0.15855)⁰ (1 - 0.15855)¹⁰⁻⁰ - (¹⁰ ₁) ( 0.15855)¹ (1 - 0.15855)¹⁰⁻¹ - (¹⁰ ₂) ( 0.15855)² (1 - 0.15855)¹⁰⁻² - (¹⁰ ₃) ( 0.15855)³ (1 - 0.15855)¹⁰⁻³

= 0.05975

Therefore the probability that more than 3 of them will have more than one ounce of mustard is 0.05975

location: point 0

step-by-step explanation:

well, -8 to 10 on a number line is 18 spaces.

1/9 of 18 is 2.

so, 4/9 of 18 must be 8.

8 spaces from -8 (a) is 0.

so, 4/9 of the way from -8 to 10 is 0!

answer:

3 bags of apples and 2 bags of oranges.

step-by-step explanation: