Answer:
a) P(x≥6)=0.633
b) P(4≤x≤8)=0.8989 (one standard deviation from the mean).
c) P(x≤7)=0.8328
Step-by-step explanation:
a) We can model this a binomial experiment. The probability of success p is the proportion of customers that prefer the oversize version (p=0.60).
The number of trials is n=10, as they select 10 randomly customers.
We have to calculate the probability that at least 6 out of 10 prefer the oversize version.
This can be calculated using the binomial expression:

b) We first have to calculate the standard deviation from the mean of the binomial distribution. This is expressed as:

The mean of this distribution is:

As this is a discrete distribution, we have to use integer values for the random variable. We will approximate both values for the bound of the interval.

The probability of having between 4 and 8 customers choosing the oversize version is:

c. The probability that all of the next ten customers who want this racket can get the version they want from current stock means that at most 7 customers pick the oversize version.
Then, we have to calculate P(x≤7). We will, for simplicity, calculate this probability substracting P(x>7) from 1.
