I need help with some 7th grade work again I BELIEVE IN YOU

Suppose it is known that 60% of radio listeners at a particular college are smokers. A sample of 500 students from the college i

vladimir1956 [14]

Answer:

The probability that at least 280 of these students are smokers is 0.9664.

Step-by-step explanation:

Let the random variable X be defined as the number of students at a particular college who are smokers

The random variable X follows a Binomial distribution with parameters n = 500 and p = 0.60.

But the sample selected is too large and the probability of success is close to 0.50.

So a Normal approximation to binomial can be applied to approximate the distribution of X if the following conditions are satisfied:

1. np ≥ 10

2. n(1 - p) ≥ 10

Check the conditions as follows:

$np=500\times 0.60=300>10\\n(1-p)=500\times(1-0.60)=200>10$

Thus, a Normal approximation to binomial can be applied.

So,

$X\sim N(\mu=600, \sigma=\sqrt{120})$

Compute the probability that at least 280 of these students are smokers as follows:

Apply continuity correction:

P (X ≥ 280) = P (X > 280 + 0.50)

= P (X > 280.50)

$=P(\frac{X-\mu}{\sigma}>\frac{280-300}{\sqrt{120}}\\=P(Z>-1.83)\\=P(Z$

*Use a z-table for the probability.

Thus, the probability that at least 280 of these students are smokers is 0.9664.

8 0

3 years ago

The figure below is dilated with the center of dilation at the origin and a scale factor

almond37 [142]

Old coordinate

B(2,-6)

As center of dilation is origin the scale factor only multiplied with B

So

new coordinate

B'

1/2(2,-6)
(2/2,-6/2)
(1,-3)

6 0

2 years ago

Solve pls brainliest! the first one to answer it gets the brainliest

mart [117]

Answer:

2.4

Step-by-step explanation:

5 0

3 years ago

Is the function represented by the table linear

Bess [88]

Answer:

No, because it does not have a constant rate of change.

Step-by-step explanation:

On the x side of the table, there is a constant rate of change (+1). However, on the y side, it is not. The first change is +7, the next +5, and the last +6. The rate of change has to be constant on both sides for the table to be considered linear.

4 0

4 years ago

A random sample of 100 people from City A has an average IQ of 120 with a SD of 18. Independently of this, a random sample of 15

Sloan [31]

Answer:

$z=\frac{(120-116)-0}{\sqrt{\frac{18^2}{100}+\frac{15^2}{150}}}}=1.837$

$p_v =P(z>1.837)=1-P(Z$

Comparing the p value with a significance level for example $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can say that the average IQ on city A is signficantly higher than city B at 5% of singificance.

Step-by-step explanation:

$\bar X_{A}=120$ represent the mean for sample 1

$\bar X_{B}=116$ represent the mean for sample 2

$s_{A}=18$ represent the sample standard deviation for 1

$s_{B}=15$ represent the sample standard deviation for 2

$n_{A}=100$ sample size for the group 2

$n_{B}=150$ sample size for the group 2

$\alpha$ Significance level provided

z would represent the statistic (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to check if residents of City A smarter on average, the system of hypothesis would be:

Null hypothesis: $\mu_{A}-\mu_{B}\leq 0$

Alternative hypothesis: $\mu_{A} - \mu_{B}> 0$

We don't have the population standard deviation's, but the sample sizes are large enough we can apply a z test to compare means, and the statistic is given by:

$z=\frac{(\bar X_{A}-\bar X_{B})-\Delta}{\sqrt{\frac{s^2_{A}}{n_{A}}+\frac{s^2_{B}}{n_{B}}}}$ (1)

z-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

With the info given we can replace in formula (1) like this:

$z=\frac{(120-116)-0}{\sqrt{\frac{18^2}{100}+\frac{15^2}{150}}}}=1.837$

P value

Since is a one right tailed test the p value would be:

$p_v =P(z>1.837)=1-P(Z$

Comparing the p value with a significance level for example $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can say that the average IQ on city A is signficantly higher than city B at 5% of singificance.

5 0

3 years ago