A recent survey found that $72 \%$ of all adults over 50 own cell phones. You randomly select 100 adults over 50 , and ask if he or she owns a cell phone.
Find the shape, mean and standard deviation of the distribution of the binomial random variable $X$.
Mean: adults over 50 .
Standard Deviation (round 2 decimals): adults over 50.
Shape: Since $n p(1-p)=$ the shape is
Using the normal approximation to the binomial, approximate the probability that at least 65 households are wireless-only households. Round to 4 decimals:
Answer
Final Answer: The shape of the distribution is approximately normal. The mean of the distribution is \(\boxed{72}\) and the standard deviation is \(\boxed{4.49}\). The probability that at least 65 households are wireless-only households is approximately \(\boxed{0.2514}\).
Step 1 ：The problem is asking for the shape, mean, and standard deviation of a binomial distribution, and then to approximate a probability using the normal approximation to the binomial distribution.
Step 2 ：The binomial distribution is defined by two parameters: the number of trials (n) and the probability of success on each trial (p). In this case, n = 100 (the number of adults over 50 that we are surveying) and p = 0.72 (the probability that an adult over 50 owns a cell phone).
Step 3 ：The mean of a binomial distribution is given by np, and the standard deviation is given by the square root of np(1-p).
Step 4 ：The shape of a binomial distribution can be approximated by a normal distribution if np and n(1-p) are both greater than 5, which they are in this case.
Step 5 ：Finally, to approximate the probability that at least 65 households are wireless-only households, we can use the normal approximation to the binomial distribution. This involves standardizing the variable (subtracting the mean and dividing by the standard deviation), and then looking up the result in a standard normal distribution table.
Step 6 ：The mean and standard deviation calculated are correct. The mean is \(\boxed{72}\) and the standard deviation is approximately \(\boxed{4.49}\).
Step 7 ：The z-score for 65 is approximately -1.56, which corresponds to a probability of 0.7486 in the standard normal distribution table. However, since we want the probability of at least 65, we need to subtract this value from 1, which gives us a probability of approximately 0.2514.
Step 8 ：Final Answer: The shape of the distribution is approximately normal. The mean of the distribution is \(\boxed{72}\) and the standard deviation is \(\boxed{4.49}\). The probability that at least 65 households are wireless-only households is approximately \(\boxed{0.2514}\).
