Suppose that birth weights are normally distributed with a mean of 3466 grams and a standard deviation of 546 grams. Babies weig
1 answer:
Answer:
3.84% probability that it has a low birth weight
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean and standard deviation
, the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
If we randomly select a baby, what is the probability that it has a low birth weight?
This is the pvalue of Z when X = 2500. So
has a pvalue of 0.0384
3.84% probability that it has a low birth weight
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Answer:
(i) The estimated regression equation is;
≈ 1.6896 + 0.0604·X
The coefficient of 'X' indicates that increase by a multiple of 0.0604 for each million dollar increase in sales, X
(ii) The estimated earnings for the company is approximately $4.7096 million
(iii) The standard error of estimate is approximately 29.34
The high standard error of estimate indicates that individual mean do not accurately represent the population mean
(iv) The coefficient of determination is approximately 0.57925
The coefficient of determination indicates that the probability of the coordinate of a new point of data to be located on the line is 0.57925
Step-by-step explanation:
The given data is presented as follows;
(i) From the data, we have;
The regression equation can be presented as follows;
= b₀ + b₁·x
Where;
b₁ = The slope given as follows;
b₀ = - b₁·
From the data, we have;
= 364.05
= 6,027.259
= 4.2
= 41.5625
∴ b₁ = 364.05/6,027.259 ≈ 0.06040059005
b₀ = 4.2 - 0.06040059005 × 41.5625 ≈ 1.68960047605 ≈ 1.69
Therefore, we have the regression equation as follows;
≈ 1.6896 + 0.0604·X
The coefficient of 'X' indicates that the earnings increase by a multiple of 0.0604 for each million dollar increase in sales
(ii) For the small company, we have;
X = $50.0 million, therefore, we get;
= 1.6896 + 0.0604 × 50 = 4.7096
The estimated earnings for the company, = 4.7096 million
(iii) The standard error of estimate, σ, is given by the following formula;
Where;
n = The sample size
Therefore, we have;
The standard error of estimate, σ ≈ 29.34
The high standard error of estimate indicates that it is very unlikely that a given mean value within the data is a representation of the true population mean
(iv) The coefficient of determination (R Square) is given as follows;
Where;
SSR = The Sum of Squared Regression ≈ 21.9884
SST = The total variation in the sample ≈ 37.96
Therefore, R² ≈ 21.9884/37.96 ≈ 0.57925
The coefficient of determination, R² ≈ 0.57925.
Therefore, by the coefficient of determination, the likelihood of a new introduced data point to located on the line is 0.57925