Zeros do not count as a sigfig so when you do something like this problem just include the numbers until you hit your significant figure limit and then substitute zeros where you can. 20000

8 0

2 years ago

A study showed that 25% of the students drive themselves to school. Based on the suggested probability, in a class of 18 student

Gwar [14]

Answer:

B) 0.283

Step-by-step explanation:

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

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

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

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

And p is the probability of X happening.

25% of the students drive themselves to school.

This means that $p = 0.25$

Class of 18 students

This means that $n = 18$

What would be the probability that at least 6 students drive themselves to school?

This is

$P(X \geq 6) = 1 - P(X < 6)$

In which

$P(X < 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)$

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

$P(X = 0) = C_{15,0}.(0.25)^{0}.(0.75)^{18} = 0.006$

$P(X = 1) = C_{15,1}.(0.25)^{1}.(0.75)^{17} = 0.034$

$P(X = 2) = C_{15,2}.(0.25)^{2}.(0.75)^{16} = 0.096$

$P(X = 3) = C_{15,3}.(0.25)^{3}.(0.75)^{15} = 0.17$

$P(X = 4) = C_{15,4}.(0.25)^{4}.(0.75)^{14} = 0.213$

$P(X = 5) = C_{15,5}.(0.25)^{5}.(0.75)^{13} = 0.199$

$P(X < 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.006 + 0.034 + 0.096 + 0.17 + 0.213 + 0.199 = 0.718$

$P(X \geq 6) = 1 - P(X < 6) = 1 - 0.718 = 0.282$

Closest option is B, just a small rounding difference.

4 0

3 years ago

The number of potato chips in a bag is normally distributed with a mean of 75 and a standard deviation of 5. Approximately what

Oxana [17]

Approximately 95% of bags contain between 65 and 85 potato chips.

<u>Step-by-step explanation</u>:

step 1 :

Given,

The mean, M = 75

Standard deviation, SD = 5

step 2 :

From Normal distribution,

The left side of mean is 32% = M-SD, 16% = M-2SD and 2% = M-3SD

The right side of mean is 32% = M+SD, 16% = M+2SD and 2% = M+3SD

step 3 :

Calculate for left side: M-SD = 75-5 = 70 and M-2SD = 75-10 = 65

Calculate for right side: M+SD = 75+5 = 80 and M+2SD = 75+10 = 85

step 4 :

Calculate the percentage between 65 and 85:

(M-SD) + (M-2SD) + (M+SD) + (M+2SD) = 32% + 16% + 32% + 16%

= 96 % (which is approximately 95 %)

4 0

3 years ago