Problem 2: The force between a positively charged foam cup and a positively charged metal sphere is 5.2 x

Samir is an expert marksman. When he takes aim at a particular target on the shooting range, there is a 0.950.950, point, 95 pro

Vinvika [58]

Answer:

40.1% probability that he will miss at least one of them

Step-by-step explanation:

For each target, there are only two possible outcomes. Either he hits it, or he does not. The probability of hitting a target is independent of other targets. So we use the binomial probability distribution to solve this question.

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.

0.95 probaiblity of hitting a target

This means that $p = 0.95$

10 targets

This means that $n = 10$

What is the probability that he will miss at least one of them?

Either he hits all the targets, or he misses at least one of them. The sum of the probabilities of these events is decimal 1. So

$P(X = 10) + P(X < 10) = 1$

We want P(X < 10). So

$P(X < 10) = 1 - P(X = 10)$

In which

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

$P(X = 10) = C_{10,10}.(0.95)^{10}.(0.05)^{0} = 0.5987$

$P(X < 10) = 1 - P(X = 10) = 1 - 0.5987 = 0.401$

40.1% probability that he will miss at least one of them

7 0

3 years ago

A researcher claims that the mean of the salaries of elementary school teachers is greater than the mean of the salaries of seco

Brut [27]

Answer:

$t=\frac{(48250-45630)}{\sqrt{\frac{3900^2}{26}+\frac{5530^2}{24}}}}=1.921$

$df=n_{A}+n_{B}-2=26+24-2=48$

Since is a one sided test the p value would be:

$p_v =P(t_{(48)}>1.921)=0.0303$

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude that the mean for elementary school teachers is significantly higher than the mean for secondary teachers at 5% of significance

Step-by-step explanation:

Data given and notation

$\bar X_{A}=48250$ represent the mean of elementary teachers

$\bar X_{B}=45630$ represent the mean for secondary teachers

$s_{A}=3900$ represent the sample standard deviation for elementary teacher

$s_{B}=5530$ represent the sample standard deviation for secondary teachers

$n_{A}=26$ sample size selected

$n_{B}=24$ sample size selected

$\alpha=0.05$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean of the salaries of elementary school teachers is greater than the mean of the salaries of secondary school teachers, the system of hypothesis would be:

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

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

We don't know the population deviations, so for this case is better apply a t test to compare means, and the statistic is given by:

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

t-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".

Calculate the statistic

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

$t=\frac{(48250-45630)}{\sqrt{\frac{3900^2}{26}+\frac{5530^2}{24}}}}=1.921$

P-value

The first step is calculate the degrees of freedom, on this case:

$df=n_{A}+n_{B}-2=26+24-2=48$

Since is a one sided test the p value would be:

$p_v =P(t_{(48)}>1.921)=0.0303$

Conclusion

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude that the mean for elementary school teachers is significantly higher than the mean for secondary teachers at 5% of significance

5 0

3 years ago

The bag of rice measures between two tens

sergeinik [125]

This is not a Question

8 0

3 years ago

Read 2 more answers

John and Alan have a collection of x baseball cards. John has x/4 cards. What fraction of the cards does Alan have?

Furkat [3]

The fraction of cards that Alan has is 3x/4. The correct answer is B.

x-x/4
4x-x/4
3x/4
he has 3x/4 cards

3 0

3 years ago

Read 2 more answers

Kamal and Anand each Lent the same sum of money for 2 years at 5 percent at simple interest and compound interest respectively.

Ivenika [448]

Answer:

amount lent: ₹6000
interest received: Kamal, ₹600; Anand, ₹615.

Step-by-step explanation:

For principal P invested at simple interest rate r, the returned value in t years is ...

A = P(1 +rt)

If K is Kamal's returned value, the given numbers tell us ...

K = P(1 +0.05·2) = 1.1P

__

For principal P invested at compound interest rate r, with interest compounded annually for t years, the returned value is ...

A = P(1 +r)^t

If A is Anand's returned value, the given numbers tell us ...

A = P(1.05)² = 1.1025P

This latter amount is RS.15 more than the former one, so we have ...

1.1025P = 1.1P +15

0.0025P = 15 . . . . . . . . subtract 1.1P

P = 6000 . . . . . . . . . . . divide by 0.0025 . . . . the amount lent

Kamal received 1.1P -P = 0.1P = 600 on the investment.

Each lent ₹6000. Kamal received ₹600 in interest; Anand received ₹615 in interest.

4 0

3 years ago