Answer:
Step-by-step explanation:
Sample mean x bar = 98.9
Std dev s = 0.62
Sample size n = 106
Std error = 
Since population std deviation not known we can use t critical value.
t critical for 99% with df = 105 is 1.98
Margin of error = ±1.98(0.0602) =±0.1192
Confidence interval for sample mean = (98.9±0.1192)
=(98.781, 99.019)
2) Sample mean confidence interval suggests that sample mean is different from the population since 98.6 is not contained in the confidence interval.
3) Population interval confidence mean would be
98.6±0.1192
=(98.4808, 98.7192)
Let Xb be the number of reservations that are accommodated. Xb has the binomial distribution with n trials and success probability p = 0.94
In general, if X has the binomial distribution with n trials and a success probability of p then
P[Xb = x] = n!/(x!(n-x)!) * p^x * (1-p)^(n-x)
for values of x = 0, 1, 2, ..., n
P[Xb = x] = 0 for any other value of x.
To use the normal approximation to the binomial you must first validate that you have more than 10 expected successes and 10 expected failures. In other words, you need to have n * p > 10 and n * (1-p) > 10.
Some authors will say you only need 5 expected successes and 5 expected failures to use this approximation. If you are working towards the center of the distribution then this condition should be sufficient. However, the approximations in the tails of the distribution will be weaker espeically if the success probability is low or high. Using 10 expected successes and 10 expected failures is a more conservative approach but will allow for better approximations especially when p is small or p is large.
If Xb ~ Binomial(n, p) then we can approximate probabilities using the normal distribution where Xn is normal with mean μ = n * p, variance σ² = n * p * (1-p), and standard deviation σ
I have noted two different notations for the Normal distribution, one using the variance and one using the standard deviation. In most textbooks and in most of the literature, the parameters used to denote the Normal distribution are the mean and the variance. In most software programs, the standard notation is to use the mean and the standard deviation.
The probabilities are approximated using a continuity correction. We need to use a continuity correction because we are estimating discrete probabilities with a continuous distribution. The best way to make sure you use the correct continuity correction is to draw out a small histogram of the binomial distribution and shade in the values you need. The continuity correction accounts for the area of the boxes that would be missing or would be extra under the normal curve.
P( Xb < x) ≈ P( Xn < (x - 0.5) )
P( Xb > x) ≈ P( Xn > (x + 0.5) )
P( Xb ≤ x) ≈ P( Xn ≤ (x + 0.5) )
P( Xb ≥ x) ≈ P( Xn ≥ (x - 0.5) )
P( Xb = x) ≈ P( (x - 0.5) < Xn < (x + 0.5) )
P( a ≤ Xb ≤ b ) ≈ P( (a - 0.5) < Xn < (b + 0.5) )
P( a ≤ Xb < b ) ≈ P( (a - 0.5) < Xn < (b - 0.5) )
P( a < Xb ≤ b ) ≈ P( (a + 0.5) < Xn < (b + 0.5) )
P( a < Xb < b ) ≈ P( (a + 0.5) < Xn < (b - 0.5) )
In the work that follows X has the binomial distribution, Xn has the normal distribution and Z has the standard normal distribution.
Remember that for any normal random variable Xn, you can transform it into standard units via: Z = (Xn - μ ) / σ
In this question Xn ~ Normal(μ = 0.94 , σ = sqrt(0.94 * n * 0.06) )
Find n such that:
P(Xb ≤ 160) ≥ 0.95
approximate using the Normal distribution
P(Xn ≤ 160.5) ≥ 0.95
P( Z ≤ (160.5 - 0.94 * n) / sqrt(0.94 * n * 0.06)) ≥ 0.95
P( Z < 1.96 ) ≥ 0.95
so solve this equation for n
(160.5 - 0.94 * n) / sqrt(0.94 * n * 0.06) = 1.96
n = 164.396
n must be integer valued so take the ceiling and you have:
n = 165.
The air line can sell 165 tickets for the flight and accommodate all reservates at least 95% of the time
If you can understand that...
Answer:
Place the dots on the x and y axis.
Step-by-step explanation:
(x, y)
Place the x axis dot on the lines that go left to right.
Place the y axis dot on the lines that go up and down.
I hope this helped! ^^
Answer:
it is not linear
Step-by-step explanation:
Linear function means that same increments in x yield uniform increments in y.
First 2 lines: x grows from -5 to 0, so increment in x is 5. For said interval, y grows from 3 to 7, so increment in y is 4.
Then, x grows from 0 to 5, so increment in x is again 5. Now, for the same increase of x, y grows from 7 to 10, so increment now in y is 3.
In the first "5" interval of x, y grew 4, and in the second "5" interval of x y grew 3 ⇒ it is not linear.
The function would have been linear, if last y value had been 11.
