Using the binomial distribution, it is found that there is a 0.5601 = 56.01% probability that the number having at least one VCR is no more than 8 but at least 6.00.

For each household, there are only two possible outcomes. Either it has at least one VCR, or it does not. The probability of a household having at least one VCR is independent of any other household, which means that the binomial distribution is used to solve this question.

Binomial probability distribution

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$C_{n,x} = \frac{n!}{x!(n-x)!}$

The parameters are:

x is the number of successes.
n is the number of trials.
p is the probability of a success on a single trial.

In this problem:

14 households, hence $n = 14$ .
0.535 probability of having at least one VCR, hence $p = 0.535$ .

The probability of <u>at least 6 and no more than 8</u> is:

$P(6 \leq X \leq 8) = P(X = 6) + P(X = 7) + P(X = 8)$

In which:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 6) = C_{14,6}.(0.535)^{6}.(0.465)^{8} = 0.1539$

$P(X = 7) = C_{14,7}.(0.535)^{7}.(0.465)^{7} = 0.2024$

$P(X = 8) = C_{14,8}.(0.535)^{8}.(0.465)^{6} = 0.2038$

Then:

$P(6 \leq X \leq 8) = P(X = 6) + P(X = 7) + P(X = 8) = 0.1539 + 0.2024 + 0.2038 = 0.5601$

0.5601 = 56.01% probability that the number having at least one VCR is no more than 8 but at least 6.00.

A similar problem is given at brainly.com/question/24863377

8 0

3 years ago

1.4t-0.4(t-3.1)=5.8<br> Answer as a decimal

zaharov [31]

it's simplified to T = 4.56 dawg

8 0

4 years ago

WRITE AN SLOPE INTERCEPT FORM

Tom [10]

1) y = x

2) y = -(1/5)x + 4

3) y = -6x + 2

4) y = x + 2

5) y = (1/2)x + 2

6) y = -x + 4

7) y = -x + 1

8) y = (3/2)x + 2

9) y = -(3/2)x - 2

10) y = 2x - 1

6 0

3 years ago

HELP ME PUT THESE IN ORDER TO LEAST TO GREATEST PLS ( i will give breanliest) HELP PLS

gregori [183]

Answer: $\sqrt{40}$ , $7\frac{1}{5}$ , $3\sqrt{6}$ , 15/2, 7.5