Answer:
a

b

Step-by-step explanation:
Let the random X variable representing the 6 companies that give 4 weeks of vacation after 15 years of employment:
-let p=0.5 be the probability of vacation. Since the companies are independent, X assumes a binomial random variable:
#Probability that the number of companies that give vacation is anywhere from 2 to 5:
We use equation 1;

Hence the probability that between 2 and 5 companies give vacation is 0.875
b. The probability that fewer than 3 companies give vacation is calculated as:
From equation one we get:

Hence the probability that less than three companies give vacation is 0.3432
Answer:

Step-by-step explanation:
We are given the following in the question:
Sample size, n = 67
Variance = 3.85
We have to find 80% confidence interval for the population variance of the weights.
Degree of freedom = 67 - 1 = 66
Level of significance = 0.2
Chi square critical value for lower tail =

Chi square critical value for upper tail =

80% confidence interval:

Putting values, we get,

Thus, (3.13,4.91) is the required 80% confidence interval for the population variance of the weights.
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