Answer:
15 scores are less than 90
Step-by-step explanation:
Total 11 scores are less than 90 not 15
To estimate, you should round some of the numbers to the closest number (rounding up or down). It will be more accurate the less you round. In this problem, we should round -5 1/4 to -5 since it is the closest. -5*-3/2 = 7.7. You will end up with -3/2x (-5 1/4) ≈ 7.5x
≈ means approximately equal.
10% of 65
Of means multiply
=0.1 x 65
=6.5
Or you can use another way which is (part/whole) *(percent/100)
Whole=65
Percent=10
And part(part of 65)=unknown
(X/65)* (10/100)
Cross multiply
x= 6.5
So 10% of 65 is 6.5
Answer:
a)

For any integer k between 0 and 15, and 0 for other values of k.
b)

c) P(6 ≤ X ≤ 10) = 0.2737
d) μ = 15*0.75 = 11.25. σ² = 11.25*0.25 = 2.8125
Step-by-step explanation:
X is a binomial random variable with parameters n = 15, p = 0.75. Therefore
a)

For any integer k between 0 and 15, and 0 for other values of k.
b)
P(X>10) = P(X=11) + P(X=12)+ P(X=13)+P(X=14)+P(x=15)





Thus,

c) P(6 ≤ X ≤ 10) = P(X = 6) + P(X = 7) + P(X = 8) + P(X=9) + P(X=10)





Thereofre,

d) μ = n*p = 15*0.75 = 11.25
σ² = np(1-p) = 11.25*0.25 = 2.8125