A cake shop bakes a variety of brownies. The top-selling brownies are ones with toppings of chocolate chip, walnuts, or both. A
customer enters the store. The probability that the customer will pick both toppings is 0.4. What is the probability that they will pick neither the chocolate chip nor the walnut toppings?
2 answers:
You might be interested in
Answer:
Using the normal distribution and the central limit theorem, it is found that there is a 0.0284 = 2.84% probability of finding a sample mean mass of 695g or below.
----------------------------------
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 establishes 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 700g means that
- Standard deviation of 21g means that
- Sample of 64, thus
- <u>For the sampling distribution of the sample mean</u>, the standard deviation is of
The probability of finding a sample mean mass of 695g or below is the p-value of Z when X = 695, thus:
By the Central Limit Theorem
has a p-value of 0.0284.
0.0284 = 2.84% probability of finding a sample mean mass of 695g or below.
A similar problem is given at brainly.com/question/22934264
Answer:
1) 22.66%
2) 20
Step-by-step explanation:
The scores of a test are normally distributed.
Mean of the test scores = u = 22
Standard Deviation = = 4
Part 1) Proportion of students who scored atleast 25 points
Since, the test scores are normally distributed we can use z scores to find this proportion.
We need to find proportion of students with atleast 25 scores. In other words we can write, we have to find:
P(X ≥ 25)
We can convert this value to z score and use z table to find the required proportion.
The formula to calculate the z score is:
Using the values, we get:
So,
P(X ≥ 25) is equivalent to P(z ≥ 0.75)
Using the z table we can find the probability of z score being greater than or equal to 0.75, which comes out to be 0.2266
Since,
P(X ≥ 25) = P(z ≥ 0.75), we can conclude:
The proportion of students with atleast 25 points on the test is 0.2266 or 22.66%
Part 2) 31st percentile of the test scores
31st percentile means 31%(0.31) of the students have scores less than this value.
This question can also be done using z score. We can find the z score representing the 31st percentile for a normal distribution and then convert that z score to equivalent test score.
Using the z table, the z score for 31st percentile comes out to be:
z = -0.496
Now, we have the z scores, we can use this in the formula to calculate the value of x, the equivalent points on the test scores.
Using the values, we get:
Thus, a test score of 20 represent the 31st percentile of the distribution.
And
(16x4-9)(x2-5)2 (16x4-25)(x4-9)
(16x4-9)(x2-5)2 (16x4-25)(x4-9)