Busn1009 Quantitative Methods For The Assessment Answer
Answer:
30% (p = 0.3) students are enrolled in marketing.
18% (p = 0.18) students are enrolled in international business.
20 students (n=20) are chosen at random.
The probability that at least half of the students (p = 0.5) are studying a marketing major = ) = P (
= (1-0.9632) = 0.0368.
The probability that no more than a quarter of the students (p = 0.75) are studying an international business major =
) = P (
= (1- 1.967*10-9).
The probability that the number of students studying a marketing major is between 10 (p = = 0.5) and 15 (p = = 0.75) is given as-
= P (
= = (0.03682 - 1.679* 10-6) = 0.036818.
The mean of the number of students studying marketing major = n*p = 20*0.3 = 6.
The standard deviation of the student studying marketing major = n*p*(1-p) = 20*0.3*(1-0.3) = 20*0.3*0.7 = 4.2 ≈ 4.
The total proportion of students studying marketing major or international business major is ()= (0.3+0.18) = 0.48.
The mean of the number of students studying marketing major = n*.
The standard deviation of the number of students studying marketing major = n**(1-) = 20*0.48*(1-0.48) = 20*0.48*0.52 = 4.99 ≈ 5.
The weights of eggs are normally distributed with average 55 gm. and standard deviation 1 gm.
The empirical rule of normality defines that the 95% confidence intervals are estimated by two standard deviations. The mean of eggs lies in the interval of –
((55 gm. - 2*1 gm.), (55 gm. + 2*1 gm.)) = ((55-2) gm., (55+2) gm.) = (53 gm., 57 gm.).
Therefore, a likely chosen random egg from the farm would have the following weight of 53 gm.
The calculated z-statistic = = 0.54577464788.
The probability for the normal distribution = P (
The calculated z-statistic = = 2.3333.
The probability for the normal distribution = P (
The probability for the normal distribution =
P (
The probability for the normal distribution =
P (
The calculated z-statistic = = -0.45977011494.
The probability for the normal distribution = P (
The calculated z-statistic = = 0.12280701754.
The probability for the normal distribution = P (
The age of real-estate investors is normally distributed that has mean (μ) = 40 years and standard deviation = 10 years.
The proportion of investors who are below 25 years old-
= P ( = P (P (
Hence, the age of 6.681% of the real-estate investors whose age is normally distributed are less than 25 years old.
The mean of home mortgage in New Zealand = $283000
The standard deviation of home mortgage in New Zealand = $50000
The home mortgages in New Zealand is normally distributed.
The proportion of home loans that is more than $250000 is given as-
P ( = P (P (
The proportion of home loans that are between $250000 and $300000 =
P ( = P (x < 300000) – P (x ≤ 250000)
= P (P (
The standard deviation = 12.56.
71.97% of the values are greater than 56.
) =
Or, = -0.582.
Or, -0.582
Or, -0.582
Or, -0.582
Or, 56 – μ = (-0.582) * (12.56) = -7.30992
Or, μ = 56 + 7.30992 = 63.30992 ≈ 63.31
Therefore, the value of μ is 63.31.
The mean = 352.
Only 13.35% of the values are less than 300.
) =
Or,
Or, = -1.11
Or,
Or,
Hence, the value of σ is 46.84685.
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