# The mean of a certain number of observations is 40. If two items

## Answers

v = 0.9875 c

Explanation:

This is an exercise of time dilation in special relativity, which is described by the equation

t = [tex]t_{p}[/tex] γ

γ = 1 / √ (1 — v² / c²)

The most important is to find respect for which system the measurement is performed.

Since the particle is still, the measured time is the proper time [tex]t_{p}[/tex] = 2.7196 10-6 s, in the system where the particles move t = 17,258 10-6 s

Let's clear dam

γ = t / [tex]t_{p}[/tex]

γ = 17,258 10-6 / 2.7196 10-6

γ = 6.345

γ = 1 / √ (1-β²)

1-β² = 1 /γ²

β = √ (1- 1 / γ²)

β = √ (1 - 1 / 6,345²)

β = 0.9875

v / c = 0.9875

v = 0.9875 c

P (3 or fewer) =0.2650

Step-by-step explanation:

Mean = x` = 5

The Poisson distribution formula is given by

P(X) = e-ˣ` x`ˣ/ x!

The mean is 5 and the X takes the values 0,1,2,and 3 which means 3 or fewer, so we add the probability of all the values of X to get the desired Value of X.

P(3 or fewer ) = e-⁵ (5)³/3!+ e-⁵ (5)²/2! +e-⁵ (5)/1!+e-⁵ (5)⁰/0!

Putting the Values

P (3 or fewer) = 0.006737 . 125 / 6 + 0.006737 . 25 / 2 +0.006737 . 5 / 1 + 0.006737 . 1 / 1

P (3 or fewer) = 0.140374 + 0.08422+ 0.03369 +0.006737

P (3 or fewer) =0.2650

Step-by-step explanation:

(a)

From the given information; we can compute the null and the alternative hypothesis as follows:

[tex]H_o : \mu = 38[/tex]

[tex]H_1 : \mu < 38[/tex]

Level of significance ∝ = 1% = 0.01

The critical values of t distribution since the sample size n = 20 is:

n - 1

= 20 - 1

= 19 degree of freedom

Assuming the population is normally distributed:

The t test can be computed by using the EXCEL FUNCTION

= TINV(0.01, 19 )

= 2.539483

[tex]t_{0.01,19} = 2.539483[/tex]

However;

we were also given the sample mean X to be = 35 minutes

the standard deviation SD = 5 minutes

Thus; the test statistics can be computed as;

[tex]t = \dfrac{\bar X - \mu}{\dfrac{s}{\sqrt{n}}}[/tex]

[tex]t = \dfrac{35- 38}{\dfrac{5}{\sqrt{20}}}[/tex]

[tex]t = \dfrac{-3}{\dfrac{5}{4.472}}[/tex]

[tex]t_o = -2.6833[/tex]

The P-value P ( t < [tex]t_o[/tex]) = P( t < - 2.633)

= 0.007355

P-value [tex]\approx[/tex] 0.0074

Decision Rule: If P - value is less than the level of significance; we are to reject the null hypothesis.

Conclusion: P-value < level of significance ; i.e 0.0074 < 0.01; so we reject the null hypothesis and accept the alternative hypothesis.

Thus; we conclude that the average time for assembling the computer board is less than 38 minutes at 0.01 level of significance.

b).

Given that:

Sample size n = 20

level of significance = 0.05

The population variance σ² is more than 22

Thus null hypothesis and the alternative hypothesis can be computed as follows:

[tex]H_0 : \sigma^2 = 22[/tex]

[tex]H_1 : \sigma^2 < 22[/tex]

From above;

degree of freedom df = 19

The critical value of [tex]X^2[/tex] at df = 19 and ∝ = 0.05 is = 30.14353 at the right tailed region.

[tex]X^2_{0.05,19} =[/tex]30.14353

The test statistics [tex]X^2[/tex] for the sample variance is computed as:

[tex]X^2= \dfrac{(n-1 )s^2}{\sigma^2}[/tex]

[tex]X^2= \dfrac{(20-1 )25}{22}[/tex]

[tex]X^2= \dfrac{(19)25}{22}[/tex]

[tex]X^2= \dfrac{475}{22}[/tex]

[tex]X^2[/tex] = 21.5909

The P-value for the test statistics is :

= 1 - P( [tex]X^2[/tex] < 21.5909)

= 1 - 0.694914

= 0.305086

The P-value = 0.305086

Decision Rule: If P - value is less than the level of significance; we are to reject the null hypothesis.

Conclusion: SInce the P-value is greater than the level of significance ; i.e

0.305086 > 0.05 ; Therefore; we do not reject the null hypothesis.

Therefore the data does not have sufficient information to conclude that the population variance is more than 22 at 5% level of significance.

I hope that helps a lot.

Step-by-step explanation:

Given that a certain population of observations is normally distributed. What percentage of the observations in the population

WE can use standard normal distribution table to get the required probabilities and hence percent.

a)P(within 1.5 std dev from the mean) = P(|z|<1.5) =2(0.4332)

=0.8664 =86.64%

b) P(more than 2.5 std dev above the mean) = P(Z>2.5) = 0.5-0.4938

=0.0062=0.62%

c) P(More than 3.5 std dev away from above or below the mean)

=P(|z|>3.5) <2(0.5-0.4990) = 0.0020

i.e. <0.2%

For the standard normal distribution Z, the mean μ =0

and σ the standard deviation = 1

a) P(0 ≤ Z ≤ 1.5) →P(Z=1.5) - P(Z=0)

P(0 ≤ Z ≤ 1.5)= 0.9332 - 0.5 = 0.4322

b) P(Z ≥ 2.5) →P(Z=2.5) = 1- P(Z=2.5)

P(Z ≥ 2.5) = 1- 0.9938 = 0.0062

c) P(Z≥ 3.5) = 1- P(Z = 3.5) = 0

OR P(Z≤ 3.5) = 0

are 20

Step-by-step explanation: