Question

In a study of 250,178 cell phone users,it was found that 81 developed cancer of the brain or nervous system. Assuming that cell phones have no effect, there is 0.000412 probability of a person developing cancer of the brain or nervous system. We therefore expect about 104 cases of such cancer in a group of 250,178 people.Estimate the probability of 81 or fewer cases of such cancer in a group of 250,178. What do these results suggest about media reports that all cell phone cause cancer of the brain or nervous system?
P(x≤81)=$P(x\leq 81)\\What do these results suggest about media reports?$

Answer 1

Using the normal distribution, the probability is given as follows:

P(x≤81) = 0.0158.

Since this probability is very small, less than 0.05, it suggest that the media reports overstated the proportion of cellphone users that developed cancer.

Normal Probability Distribution

The z-score of a measure X of a normally distributed variable with mean $\mu$ and standard deviation $\sigma$ is given by:

$Z = \frac{X - \mu}{\sigma}$

• The z-score measures how many standard deviations the measure is above or below the mean.

• Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

• The binomial distribution is the probability of x successes on n trials, with p probability of a success on each trial. It can be approximated to the normal distribution with $\mu = np, \sigma = \sqrt{np(1-p)}$.

For this problem, the proportion and the sample size are given by:

p = 0.000412, n = 250178

Hence the mean and the standard deviation for the approximation are given by:

• $\mu = np = 250178 \times 0.000412 = 103.1$.

• $\sigma = \sqrt{np(1-p)} = \sqrt{250178 \times 0.000412 \times (1 - 0.000412)} = 10.15$

The probability that x is of 81 or less than the p-value of Z when X = 81.5, accounting for continuity correction, hence:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{81.5 - 103.1}{10.15}$

Z = -2.13

Z = -2.13 has a p-value of 0.0158.

Hence:

P(x≤81) = 0.0158.

Since this probability is very small, less than 0.05, it suggest that the media reports overstated the proportion of cellphone users that developed cancer.

More can be learned about the normal distribution at https://brainly.com/question/4079902

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