I need help with number 15,16,and 17

Calculate the semi-interquartile range of the data shown 21; 34; 53, 43; 27; 38; 20; 32; 13;42:55

yKpoI14uk [10]

Answer:

21yrnrndjen edjdje ebsjnednzhdhr

4 0

3 years ago

Lisa's age is 4 years older than three times Sammy's age. Lisa is 28 years old. Let s represent Sammy's age.

elena-s [515]

3s + 4 = 28 Lisa is 4 years older than three times his age. 4 years older would mean +4 , and the 3 times his age is 3s because the s represents sammys age

4 0

4 years ago

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Two alloys A and B are being used to manufacture a certain steel product. An experiment needs to be designed to compare the two

sveta [45]

Answer:

Step-by-step explanation:

From the given information, we can compute the table showing the summarized statistics of the two alloys A & B:

Alloy A Alloy B

Sample mean $\bar {x} _A = 49.5$ $\bar {x} _B = 45.5$

Equal standard deviation $\sigma_A = 5$ $\sigma_B= 5$

Sample size $n_A = 30$ $n_{B}= 30$

Mean of the sampling distribution is :

$\mu_{\bar{X_1}-\bar{X_2}}= 49.5-45.5 \\ \\ = 4.0$

Standard deviation of sampling distribution:

$\sigma_{\bar{x_1}-\bar{x_2} }= \sqrt{\sigma^2_{\bar{x_1}-\bar{x_2} }} \\ \\ = \sqrt{\dfrac{\sigma_1^2}{n_1} + \dfrac{\sigma^2_2}{n_2} } \\ \\ =\sqrt{\dfrac{25}{30}+\dfrac{25}{30}} \\\\=\sqrt{1.667} \\ \\ =1.2909$

Hypothesis testing.

Null hypothesis: $H_o: \mu_A -\mu_B = 0$

Alternative hypothesis: $H_A:\mu_A -\mu_B > 4$

The required probability is:

$P(\overline X_A - \overline X_B>4|\mu_A - \mu_B) = P\Big (\dfrac{(\overline X_A - \overline X_B)-\mu_{X_A-X_B}}{\sigma_{\overline x_A -\overline x_B}} > \dfrac{4 - \mu_{X_A-\overline X_B}}{\sigma _{\overline x_A - \overline X_B}} \Big) \\ \\ = P \Big( z > \dfrac{4-0}{1.2909}\Big) \\ \\ = P(z \ge 3.10)\\ \\ = 1 - P(z < 3.10) \\ \\ \text{Using EXCEL Function:} \\ \\ = 1 - [NORMDIST(3.10)] \\ \\ = 1- 0.999032 \\ \\ 0.000968 \\ \\ \simeq 0.0010$

This implies that a minimal chance of probability shows that the difference of 4 is not likely, provided that the two population means are the same.

b)

Since the P-value is very small which is lower than any level of significance.

Then, we reject $H_o$ and conclude that there is enough evidence to fully support alloy A.

3 0

3 years ago

100 random students are surveyed outside of the student center on campus. Let V denote that the student has a Visa credit card a

mote1985 [20]

Answer:

(a) The value of P (M | V) is 0.30.

(b) The value of $P (M^{c} | V)$ is 0.70.

(c) The value of P (V | M) is 0.375.

(d) The value of $P(V^{c}|M)$ is 0.625.

Step-by-step explanation:

It is provided that,

V = a student has a Visa card

M = a student has a Master card

N = 100, n (V) = 40, n (M) = 32 and n (V ∩ M) = 12.

The probability of a student having visa card is:

$P(V) = \frac{n(V)}{N}= \frac{40}{100}=0.40$

The probability of a student having master card is:

$P(M) = \frac{n(M)}{N}= \frac{32}{100}=0.32$

The probability of a student having visa card and a master card is:

$P(V\cap M) = \frac{n(V\cap M)}{N}= \frac{12}{100}=0.12$

The conditional probability of an event, say A, given that another event, say B, has already occurred is,

$P(A|B)=\frac{P(A\cap B)}{P(B)}$

(a)

Compute the probability that a student has a master card given that he/she has a visa card also, i.e. P (M | V) as follows:

$P(M|V)=\frac{P(V\cap M)}{P(V)} =\frac{0.12}{0.40}=0.30$

Thus, the value of P (M | V) is 0.30.

(b)

Compute the probability that a student does not have a master card given that he/she has a visa card also, i.e. $P (M^{c} | V)$ as follows:

$P (M^{c} | V)=1-P(M|V)=1-0.30=0.70$

Thus, the value of $P (M^{c} | V)$ is 0.70.

(c)

Compute the probability that a student has a visa card given that he/she has a master card also, i.e. P (V | M) as follows:

$P(V|M)=\frac{P(V\cap M)}{P(M)} =\frac{0.12}{0.32}=0.375$

Thus, the value of P (V | M) is 0.375.

(d)

Compute the probability that a student does not have a visa card given that he/she has a master card also, i.e. $P(V^{c}|M)$ as follows:

$P(V^{c}|M)=1-P(V|M)=1-0.375=0.625$

Thus, the value of $P(V^{c}|M)$ is 0.625.

8 0

3 years ago

What is 3-6+9-7+9+9-3-7+19-6?

Ahat [919]

Answer:

answer is 20 ..........

3 0

3 years ago

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