In a population of similar households, suppose the weekly supermarket expense for a typical household is normally distributed with mean $135 and standard deviation $12. If you observe 50 households from this population, the probability that at least 15 of them have supermarket expenses of more than $140 in a given week is closest to which of the following? a. 0.763
b. 0.660
c. 0.237
d. 0.340

P(Y ≥ 15) = 0.763

Step-by-step explanation:

Given that:

Mean =135

standard deviation = 12

sample size n  = 50

sample mean = 140

Suppose X is the random variable that follows a normal distribution which represents the weekly supermarket expenses

Then,

The probability that X is greater than 140 is :

P(X>140) = 1 - P(X ≤ 140)

From z tables,

Similarly, let consider Y to be the variable that follows a binomial distribution of the no of household whose expense is greater than $140

Then;

∴

P(Y ≥ 15) = 1- P(Y< 15)

P(Y ≥ 15) = 1 - ( P(Y=0) + P(Y=1) + P(Y=2) + ... + P(Y=14) )

P(Y ≥ 15) = 0.763

the solution is the point

step-by-step explanation:

we have

> equation a

> equation b

adds equation a and equation b

substitute the value of x in equation b to find the value of y

the solution is the point

hello from mrbilldoesmath!

answer:   x (x+6)^2, choice b

discussion:

factor x out of the equation:

x^3 + 12x^2 + 36x = x (x^2 + 12x + 36).

note that 6+6 = 12 and 6*6 = 36 so the above factors into:

x (x+6)^2

which is choice b

you,

mrb