Please help me with these 6 questions

The mean weight 20 students in a room is 65 kg one student one student who weight is 88kg went into the room what is the new mea

umka2103 [35]

The new mean is 66.09

<u>Explanation:</u>

Mean weight of 20 students = 65kg

One student with 88kg entered the room.

New mean = ?

$\frac{x_1+x_2+....+x_2_0}{20} = 65$

$\frac{x_1+x_2+....+x_2_0 + 88}{21} = ?$

The above equation can further be written as:

$x_1 + x_2 +....+x_2_0 = 65 X 20\\\\x_1 + x_2 +....+x_2_0 = 1300$

New mean = $\frac{1300 + 88}{21}$

= 66.09

Therefore, the new mean is 66.09

6 0

3 years ago

Ava read 126 books in 12 months. How many books per month did she read? Show your work

Sladkaya [172]

10.5 hopes this helps

4 0

3 years ago

Is it true that for any two nonzero integers the product and quotient have the same sign?

Nataliya [291]

<span>The rule of signs is the same for multiplication and division. Two numbers with the same sign, multiplied or divided, the answer is positive. Two numbers with different signs, multiplied or divided, the answer is negative. </span>

7 0

3 years ago

At Western University the historical mean of scholarship examination scores for freshman applications is 900. A historical popul

vampirchik [111]

Answer:

a) Null Hypothesis: $\mu =900$

Alternative hypothesis: $\mu \neq 900$

b) The 95% confidence interval would be given by (910.05;959.95)

c) Since we confidence interval not ocntains the value of 900 we fail to reject the null hypothesis that the true mean is 900.

d) $z=\frac{935 -900}{\frac{180}{\sqrt{200}}}=2.750$

Since is a bilateral test the p value is given by:

$p_v =2*P(Z>2.750)=0.0059$

Step-by-step explanation:

a. State the hypotheses.

On this case we want to check the following system of hypothesis:

Null Hypothesis: $\mu =900$

Alternative hypothesis: $\mu \neq 900$

b. What is the 95% confidence interval estimate of the population mean examination score if a sample of 200 applications provided a sample mean x¯¯¯= 935?

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X=935$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

$\sigma=180$ represent the population standard deviation

n=200 represent the sample size

The confidence interval for the mean is given by the following formula:

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (1)

In order to calculate the mean and the sample deviation we can use the following formulas:

$\bar X= \sum_{i=1}^n \frac{x_i}{n}$ (2)

$s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}}$ (3)

The mean calculated for this case is $\bar X=3278.222$

The sample deviation calculated $s=97.054$

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that $z_{\alpha/2}=1.96$