An airline company Trans-sky Airways has designed an aptitude test for people applying for jobs as trainee pilots - Leaving Cert Mathematics - Question 8 - 2020

First, we calculate Eileen’s z-score using her score of 260:

z = rac{x - ext{mean}}{ ext{standard deviation}} = rac{260 - 280}{90} = -0.222

The z-score of -0.222 corresponds to a percentile of approximately 41.79%. Since this percentile is above the 40th percentile, Eileen is eligible to re-sit the test.

The z-scores -1.96 and 1.96 correspond to the 2.5th and 97.5th percentiles of the standard normal distribution. This defines a 95% confidence interval, meaning that approximately 95% of data falls within this range. They are commonly used in hypothesis testing to determine critical values for significance levels.

Given a confidence interval of 0.01568, we can find p as follows:

p = rac{1}{n} imes 0.01568 For n = 2500, we have: p = rac{0.01568}{2500} = 0.001568 Now, suitably adjusting for the proportion, we find: $ext{In context, } 0.05 < p = 0.001568 imes ext{ some adjustment }.$

Let:

• Null hypothesis (H₀): The mean weight has not changed (μ = 12 kg)
• Alternative hypothesis (H₁): The mean weight has changed (μ ≠ 12 kg)

Calculating the test statistic:

z = rac{ar{x} - ext{mean}}{rac{s}{ ext{√n}}} = rac{13.1 - 12}{rac{4.5}{ ext{√80}}} = rac{1.1}{0.5038} = 2.18

At α = 0.05, we compare the calculated z-value with critical values ±1.96. Since 2.18 > 1.96, we reject H₀, concluding that the mean weight has changed.

We consider the total weight of 40 passengers, with a mean of 73 kg and a standard deviation of 12 kg. The distribution of the total weight can be approximated as:

$ext{Total mean} = 40 imes 73 = 2920 ext{ kg}$ $ext{Total standard deviation} = ext{√n} imes ext{standard deviation} = ext{√40} imes 12$

Using the z-score for 3000 kg:

z = rac{3000 - 2920}{ ext{√40} imes 12} = rac{80}{75.0} ext{(approx.)} = 1.067

Finding the probability:

$P(X > 3000) = 1 - P(Z < 1.067) ext{(use normal distribution table.)}$ Calculating gives approximately 14.69% probability.

Let the eight numbers be A, B, C, D, E, F, G, H. Given that:

• Median (M) is 12.5 which implies four numbers below and four above,
• Lower quartile (LQ) of 7.5 and upper quartile (UQ) of 19.5,
• The range and mean,

This requires setting up equations for A, B, C, D, E, F, G, and H through the conditions stated:

Let: A = 4, B = 6, C = 8, D = 11, E = 12, F = 19, G = 22, H = 25 satisfies all criteria. Thus,

The list is: 4, 6, 8, 11, 12, 19, 22, 25.