Find the sample variance and standard deviation of 18, 16, 5, 10, 12

It has been estimated that only about 30% of California residents have adequate earthquake supplies. Suppose you randomly survey

mote1985 [20]

Answer:

$P(X\geq 8)=0.0043\\\\$

It's more likely that all of the residents surveyed will have adequate earthquake supplies since it has a probability of 98.02% which is very close to 100%.

Step-by-step explanation:

-This is a binomial probability problem with the function:

$P(X=x)={n\choose x}p^x(1-p)^{n-x}$

-Given p=0.3, n=11, the is calculated as:

$P(X=x)={n\choose x}p^x(1-p)^{n-x}\\\\P(X\geq 8)=P(X=8)+P(X=9)+P(X=10)+P(X=11)\\\\={11\choose 8}0.3^8(0.7)^3+{11\choose 9}0.3^9(0.7)^2+{11\choose 10}0.3^{10}(0.7)^1+{11\choose 11}0.3^{11}(0.7)^0\\\\=0.0037+0.0005+0.00005+0.000002\\\\=0.0043$

Hence, the probability that at least 8 have adequate supplies 0.0043

#The probability that non has adequate supplies is calculated as;

$P(X=x)={n\choose x}p^x(1-p)^{n-x}\\\\P(X= 0)={11\choose 0}0.3^{0}(0.7)^{11}\\\\=0.0198$

#The probability that all have adequate supplies is calculated as:

$P(X=x)={n\choose x}p^x(1-p)^{n-x}\\\\P(X= All)=1-{11\choose 0}0.3^{0}(0.7)^{11}\\\\=1-0.0198\\\\=0.9802$

Hence, it's more likely that all of the residents surveyed will have adequate earthquake supplies since $P(All)>P(None)\ \$ and that this probability is 0.9802 or 98.02% a figure close to 1

7 0

3 years ago

The sum of two prime numbers is 8585. What is the product of these two prime numbers?

dybincka [34]

8585 is odd, so one of the primes is even and the other is odd (otherwise their sum would be even). 2 is the only even prime. Thus, second prime is 8585-2=8583. Their product is 8583*2=<span>17166.</span>

4 0

3 years ago

Two stores sell the same refrigerator for the same original price. Store A advertises that the refrigerator is on sale for 15% o

ololo11 [35]

The correct answer is option A. Since store A is giving a discount of 15% on the original price i.e.$ p. Then, the price after discount will be equal to $ 0.85p . And, the second store B offers $150 less than the original price p i.e the price after reduction is $p-$150. Since both the proces after reduction are equal. Thus, $ 0.85p = $p-$150

6 0

3 years ago

Taylor surveys students in one grade level who own at least one pet. She finds that 50% of the students surveyed own 2 pets, 3 s

Llana [10]

First, we need to work out the total number of students who were being surveyed.

We know that half of the students has two pets. The rest of the students make up the other half. So, we have 3 students + 2 students + 8 students = 13 students that make half of the sample population

That means total number of students being surveyed is 13+13=26 students

Then we work out the probability
P(One pet) = 8/26 = 4/13
P(Two pets) = 1/2
P(Three pets) = 3/26
P( Four pets) = 2/26 = 1/13

The probability distribution is shown in the table below. Let $x$ be the number of pets and $P(X=x)$ is the probability of owning the number of pets