50% of cakes were party cakes 1/5 were fruit cakes and remainder were sponge cakes what % were sponge cakes"
2 answers:
the total percentage is 100 %
there are party cakes, fruit cakes and sponge cakes
50 % of the cakes are party cakes
1/5th of the cakes are fruit cakes.
so 1/5th of 100 % is - = 20%
therefore 20 % of the cakes are fruit cakes
remainder of the cakes were sponge cakes
the sum of the party and fruit cakes are - 50 % + 20 % = 70 %
remainder is 100 % - 70 % = 30 %
therefore 30 % of the cakes were sponge cakes
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Using the <u>normal probability distribution and the central limit theorem</u>, it is found that:
a) There is a 0.1922 = 19.22% probability that Company B would charge more than Company A to deliver a small-item package.
b) The mean amount charged is of RM 51.5, with a standard deviation of 21.35.
In a normal distribution with mean and standard deviation
, the z-score of a measure X is given by:
- It measures how many standard deviations the measure is from the mean.
- After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.
By the Central Limit Theorem:
- When a <u>fixed constant k</u> multiplies a variable, the mean is
and the standard deviation is
- When two normal variables are subtracted, the mean is the subtraction of the means, while the standard deviation is the square root of the sum of the variances.
In this question, b is needed to solve a, so I am going to place the solution to item b first.
Item b:
Flat fee of RM20(not considered for the standard deviation), plus a variable fee of RM3.5, hence:
The mean amount charged is of RM 51.5, with a standard deviation of 21.35.
Item a:
This probability is . For the distribution, we have that:
Since A has a constant fee, it's standard deviation is 0, hence:
The probability is <u>1 subtracted by the p-value of Z when X = 0</u>, so:
has a p-value of 0.8078.
1 - 0.8078 = 0.1922.
0.1922 = 19.22% probability that Company B would charge more than Company A to deliver a small-item package.
A similar problem is given at brainly.com/question/25403659