Final answer:
Without the specific probability of finding the gene linked to a childhood disease in the general population, we cannot determine the exact z-score and therefore cannot provide an answer to whether it would be unlikely to find fewer than 25 children with the gene out of 950 newborns.
Step-by-step explanation:
To determine if it would be unlikely to find fewer than 25 children with the gene linked to a childhood disease out of 950 newborns, we can calculate the z-score. First, we need the probability (or the expected proportion) of finding the gene in the general population. This information is missing in the question, but for the sake of the example let's assume that the gene's presence in the population is p.
The mean (μ) number of children with the gene in a sample of 950 is given by μ = 950p. The standard deviation (σ) of the number of children with the gene is σ = sqrt(950p(1-p)). If we find fewer than 25 children with the gene, we will calculate the z-score using the formula z = (X - μ) / σ, where X is the number of children with the gene we are testing for - in this case, 25.
Without the actual probability (p), we cannot calculate the specific z-score, but we can say that if the z-score is less than -1.96 (the critical value for a 0.05 significance level), it would be considered statistically unlikely. If the z-score is between -1.96 and +1.96, it is within the expected range, and if it is greater than +1.96, it would be considered unusually high.
Therefore, without the exact probability, we cannot choose the correct option from a) No, z=−0.005 b) Yes, z=−4.41 c) No, z=−6.59 d) Yes, z=−6.59. If we had the probability, we could calculate the exact z-score and answer the student's question accurately.