Need help wit questions 2,3,4,

Identify the symbols used for each of the following: (a) sample standard deviation; (b) population standard deviation; (c) sa

Anvisha [2.4K]

Answer:

HEHE

Step-by-step explanation:

5 0

3 years ago

Find two consecutive even numbers whose sum is 758.

ra1l [238]

Answer:

378 and 380

Step-by-step explanation:

The two even consecutive numbers that add up to 758 are going to be very close to half of 758. This is because two half of 758 are going to be the most similar addends of 758. This is important because the answers will be consecutive and therefore, must also be very similar. To solve, first, divide 758 by 2. This is 379, which is not an even number. So, to find the needed addends subtract and add 1 to 379. Both of these will be even and consecutive. These two numbers are 378 and 380. Then, to check you, can add them and see that they do sum 758.

7 0

3 years ago

Read 2 more answers

A common practice of airline companies is to sell more tickets for a particular flight than there are seats on the plane, becau

Serga [27]

Answer:

0.0159

Step-by-step explanation:

Given that a common practice of airline companies is to sell more tickets for a particular flight than there are seats on the plane, because customers who buy tickets do not always show up for the flight.

Here if X is the no of persons that do not show up, then X is binomial as each trial is independent with p = 0.04 and n =150 (no of tickets sold)

The plane is overbooked if more than 150 show up

i.e. less than 2 do not show up

Hence the probability that the airline overbooked this flight

= $P(X$

4 0

3 years ago

Please help :) Please and thank you :D

Olin [163]

Answer I think the first one

Step-by-step explanation:Hope this helps

4 0

3 years ago

The International Air Transport Association surveys business travelers to develop quality ratings for transatlantic gateway airp

motikmotik

Answer:

The 95% confidence interval for the population mean rating is (5.73, 6.95).

Step-by-step explanation:

We start by calculating the mean and standard deviation of the sample:

$M=\dfrac{1}{n}\sum_{i=1}^n\,x_i\\\\\\M=\dfrac{1}{50}(6+4+6+. . .+6)\\\\\\M=\dfrac{317}{50}\\\\\\M=6.34\\\\\\s=\sqrt{\dfrac{1}{n-1}\sum_{i=1}^n\,(x_i-M)^2}\\\\\\s=\sqrt{\dfrac{1}{49}((6-6.34)^2+(4-6.34)^2+(6-6.34)^2+. . . +(6-6.34)^2)}\\\\\\s=\sqrt{\dfrac{229.22}{49}}\\\\\\s=\sqrt{4.68}=2.16\\\\\\$

We have to calculate a 95% confidence interval for the mean.

The population standard deviation is not known, so we have to estimate it from the sample standard deviation and use a t-students distribution to calculate the critical value.

The sample mean is M=6.34.

The sample size is N=50.

When σ is not known, s divided by the square root of N is used as an estimate of σM:

$s_M=\dfrac{s}{\sqrt{N}}=\dfrac{2.16}{\sqrt{50}}=\dfrac{2.16}{7.071}=0.305$