The sum of two numbers is 128. Their difference is 114. Find the numbers
1 answer:
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Answer:
a) P(X < 99) = 0.2033.
b) P(98 < X < 100) = 0.4525
Step-by-step explanation:
To solve this question, we need to understand the normal probability distribution and the central limit theorem.
Normal Probability Distribution:
Problems of normal distributions can be solved using the z-score formula.
In a set with mean and standard deviation
, the z-score of a measure X is given by:
The Z-score 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. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.
Central Limit Theorem
The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean and standard deviation
, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
and standard deviation
.
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
Mean of 100 and variance of 36.
This means that
Sample of 25:
This means that
(a) P(X<99)
This is the pvalue of Z when X = 99. So
By the Central Limit Theorem
has a pvalue of 0.2033. So
P(X < 99) = 0.2033.
b) P(98 < X < 100)
This is the pvalue of Z when X = 100 subtracted by the pvalue of Z when X = 98. So
X = 100
has a pvalue of 0.5
X = 98
has a pvalue of 0.0475
0.5 - 0.0475 = 0.4525
So
P(98 < X < 100) = 0.4525
Answer:
62.29%
Step-by-step explanation:
The probability of Aya being offered a coupon on at least one of the six days she visits the website is 100% minus the probability that she is not offered a coupon on any of the six days, which is described by a binomial probability with zero successes in six trials with a probability of succes p = 0.15.
The probability that Aya will be offered a coupon on at least one of the days she visits the website is 62.29%.