Ask question

Ask question

All categories

3 years ago

8

in a class of 40 students, 24 are over the age of 21 and 16 are 21 or younger. Biff is over the age of 21. A student is selected

at random. (3 pts each)3. Find the probability the student is over the age of 21.4. Find the probability Biff was selected if you know the student selected was over the age of 21. 5. Find the probability Biff was selected if you know the student selected was 21 or younger

1 answer:

LenKa [72]3 years ago

8 0

Answer:

3. 60% probability the student is over the age of 21.4

4. 4.17% probability Biff was selected if you know the student selected was over the age of 21.

5. 0% probability Biff was selected if you know the student selected was 21 or younger

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcoes.

In this question:

40 students

24 over the age of 21.

16 are 21 or younger.

3. Find the probability the student is over the age of 21.

24 of the 40 students are over the age of 21.

24/40 = 0.6

60% probability the student is over the age of 21.4

4. Find the probability Biff was selected if you know the student selected was over the age of 21.

24 students over the age of 21, one of which is Biff.

1/24 = 0.0417

4.17% probability Biff was selected if you know the student selected was over the age of 21.

5. Find the probability Biff was selected if you know the student selected was 21 or younger

Biff is not 21 or younger, he is over 21, so there are no desired outcomes.

So, 0% probability Biff was selected if you know the student selected was 21 or younger

You might be interested in

The graph of an equation drawn through which two points would best represent the relationship between the number of miles and th

andreyandreev [35.5K]

Answer:

250

Step-by-step explanation:

Plz ve me brainlest

5 0

3 years ago

What is the product of (7x+2) and 20-3x/4

fredd [130]

(7x+2)(20-3x/4) = 140x + 40 - 21x²/4 - 3x/2 = -21/4 x² + 138 1/2 x + 40

this is one of those multiplications that doesn't make the result look any simpler...

3 0

3 years ago

Help me plz plz plz plz plz plz

DENIUS [597]

Answer:

yksykskyskgsykslydhdlhdyldkydkydydyodyldgkdlydlyd

7 0

3 years ago

A recipe uses 5 cups of flour , Nadia is making 1/2 of the recipe . How many cups of flour does Nadia need ? Move numbers to the

fredd [130]

Answer:

2.5 rounds to 3

Step-by-step explanation:

5 divided by (or 5/2) 2 is 2.5 and since its not a whole number you rounded by the ones and you get 3.

3 0

3 years ago

A medical researcher wondered if there is a significant difference between the mean birth weight of boy and girl babies. Random

Yuliya22 [10]

Answer:

b) independent samples t-test

$t=\frac{(5.92 -5.74)-(0)}{\sqrt{\frac{1.88^2}{5}}+\frac{1.81^2}{5}}=0.154$

$p_v =2*P(t_{8}>0.154) =0.881$

So with the p value obtained and using the significance level given $\alpha=0.05$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the mean of the group 1 (boys) is NOT significantly different than the mean for the group 2 (Girls).

Step-by-step explanation:

The system of hypothesis on this case are:

Null hypothesis: $\mu_1 = \mu_2$

Alternative hypothesis: $\mu_1 \neq \mu_2$

Or equivalently:

Null hypothesis: $\mu_1 - \mu_2 = 0$

Alternative hypothesis: $\mu_1 -\mu_2\neq 0$

Our notation on this case :

$n_1 =5$ represent the sample size for group 1

$n_2 =5$ represent the sample size for group 2

We can calculate the mean and deviation for eaach group with the following formulas:

$\bar X= \frac{\sum_{i=1}^n X_i}{n}$

$s = \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

$\bar X_1 =5.92$ represent the sample mean for the group 1

$\bar X_2 =5.74$ represent the sample mean for the group 2

$s_1=1.88$ represent the sample standard deviation for group 1

$s_2=1.81$ represent the sample standard deviation for group 2

If we see the alternative hypothesis we see that we are conducting a bilateral test and with independnet samples t test.

b) independent samples t-test

The statistic is given by this formula:

$t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{\sqrt{\frac{s^2_1}{n_1}}+\frac{S^2_2}{n_2}}$

And now we can calculate the statistic:

$t=\frac{(5.92 -5.74)-(0)}{\sqrt{\frac{1.88^2}{5}}+\frac{1.81^2}{5}}=0.154$

The degrees of freedom are given by:

$df=5+5-2=8$

And now we can calculate the p value using the altenative hypothesis:

$p_v =2*P(t_{8}>0.154) =0.881$

So with the p value obtained and using the significance level given $\alpha=0.05$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the mean of the group 1 (boys) is NOT significantly different than the mean for the group 2 (Girls).

7 0

3 years ago