A newborn baby whose apgar score is over 6 is classified as normal and this happens in 80% of births. As a quality control check an auditor examines the records of 100 births. He would be suspicious if the number of normal births in the sample of 100 births fell above the upper limit of a 95% normal range. What is the upper limit?
Interesting problem.
Check if binomial theorem applies.
1. Is each trial a Bernoulli trial (either success or failure)? Yes
2. Is probability of success (Apgar><6) known and remain constant throughout experiment (audit)? Yes
3. Are trials (result of each birth) independent of each other? Basically yes, no indication to the contrary.
4. The number of trials is known and remains constant (100). Yes
Since all criteria are satisfied, we can use the binomial distribution to model the experiment, with which:
n=number of trials = 100
p=probability of success = 0.8
q=probability of failure = 1-0.8=0.2
μ=np = 100*0.8=80
σ²=npq=100*0.8*0.2=16
σ=√16=4
Since n>40, np>10, nq>10, we can approximate the distribution as a normal distribution (with appropriate continuity corrections).
The auditor is suspicious only if the numbers are too high, i.e. a upper-tail test.
So the limit is
μ+;1.644*σ
=86.58 Say 87.
So auditor would be suspicious if the number of normal births is above 87, for a sample of 100 births, at the 95% level.
To find the upper limit of the 95% normal range, we need to calculate the mean and standard deviation of the number of normal births in the sample.
Given that the probability of a normal birth is 80%, the mean value of the number of normal births in a sample of 100 births can be calculated as:
Mean = (sample size) x (probability of normal birth) = 100 x 0.80 = 80
The standard deviation (SD) can be calculated using the formula:
SD = √(n x p x q)
where n is the sample size, p is the probability of success (normal birth), and q is the probability of failure (1 - p).
In this case:
SD = √(100 x 0.80 x 0.20) = √(16) = 4
To find the upper limit of the 95% normal range, we need to calculate the Z-score corresponding to the 95th percentile. The Z-score is calculated as:
Z = (X - Mean) / SD
At the 95th percentile (or α = 0.05), the Z-score is approximately 1.645 based on a standard normal distribution table.
Now, we can calculate the upper limit of the 95% normal range by multiplying the Z-score by the standard deviation and adding it to the mean:
Upper Limit = Mean + (Z-score x SD)
Upper Limit = 80 + (1.645 x 4)
Upper Limit ≈ 80 + 6.58 ≈ 86.58
Therefore, the upper limit of the 95% normal range for the number of normal births in the sample is approximately 86.58.
To find the upper limit of the 95% normal range, we need to calculate the mean and standard deviation of the number of normal births in the sample.
Given that 80% of births are classified as normal, we can expect that the mean number of normal births in a sample of 100 would be 80.
To calculate the standard deviation, we need to know the variance of the number of normal births. Since we are not given this information directly, we can use the binomial distribution approximation to estimate it.
For a binomial distribution, the variance is given by the formula:
Var(X) = n * p * (1 - p)
Where:
- n is the sample size (100 in this case)
- p is the probability of success (0.8 in this case)
Substituting the values, we get:
Var(X) = 100 * 0.8 * (1 - 0.8)
Var(X) = 100 * 0.8 * 0.2
Var(X) = 16
The standard deviation (σ) is the square root of the variance:
σ = √16
σ = 4
Now we can calculate the upper limit of the 95% normal range using the Z-score formula.
A Z-score represents the number of standard deviations a particular value is from the mean. For a 95% confidence level, the Z-score is approximately 1.96 (obtained from a standard normal distribution table).
The upper limit can be calculated as:
Upper Limit = Mean + (Z-score * Standard Deviation)
Upper Limit = 80 + (1.96 * 4)
Upper Limit ≈ 80 + 7.84
Upper Limit ≈ 87.84
Therefore, the upper limit of the 95% normal range for the number of normal births in the sample of 100 births is approximately 87.84. If the number of normal births in the sample exceeds this upper limit, the auditor would be suspicious.
So the limit is
μ+;1.644*σ
=86.58 Say 87.
So auditor would be suspicious if the number of normal births is above 87, for a sample of 100 births, at the 95% level
plz explain