Answer:
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
