if 1/3 of a can of frosting covers 1/6 of the top of the birthday cake how much frosting is needed to cover the entire top of th
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Using the <u>normal approximation to the binomial</u>, it is found that there is a 0.994 = 99.4% probability that we will have enough seats for everyone who shows up.
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.
- The binomial distribution is the probability of <u>x successes on n trials</u>, with <u>p probability</u> of a success on each trial. It can be approximated to the normal distribution with
.
In this problem:
- 15% do not show up, so 100 - 15 = 85% show up, which means that
.
- 300 tickets are sold, hence
.
The mean and the standard deviation are given by:
The probability that we will have enough seats for everyone who shows up is the probability of at most <u>270 people showing up</u>, which, using continuity correction, is , which is the <u>p-value of Z when X = 270.5</u>.
has a p-value of 0.994.
0.994 = 99.4% probability that we will have enough seats for everyone who shows up.
A similar problem is given at brainly.com/question/24261244