A recent study has shown that $28 \%$ of $18-34$ year olds check their Facebook/Instagram feeds before getting out of bed in the morning.
If we sampled a group of $15018-34$ year olds, what is the probability that the number of them who checked their social media before getting out of bed is:
a.) At least 32 ?
b.) No more than 50 ?
c.) between 38 and 49 (including 38 and 49 )?
Use the Normal Approximation to the Binomial distribution to answer this question.
Question Help: MMessage instructor
Answer
Final Answer: The probability that at least 32 out of 150 18-34 year olds check their social media before getting out of bed is approximately \(\boxed{0.966}\). The probability that no more than 50 do so is approximately \(\boxed{0.927}\). The probability that between 38 and 49 (inclusive) do so is approximately \(\boxed{0.665}\).
Step 1 ：Given that the number of trials, \(n = 150\), and the probability of success, \(p = 0.28\).
Step 2 ：Calculate the mean and standard deviation of the binomial distribution. The mean of a binomial distribution is \(np\), and the standard deviation is \(\sqrt{np(1-p)}\).
Step 3 ：Substitute the given values into the formulas to get the mean and standard deviation. The mean is \(150 \times 0.28 = 42.00000000000001\) and the standard deviation is \(\sqrt{150 \times 0.28 \times (1-0.28)} = 5.499090833947008\).
Step 4 ：For each part of the question, convert the number of successes to a z-score using the formula \(z = (x - \text{mean}) / \text{standard deviation}\).
Step 5 ：For part a, the z-score is \((-1.818482418633271)\). Using a z-table or a function that gives the cumulative distribution function of the normal distribution, the probability is approximately \(0.9655047842587428\).
Step 6 ：For part b, the z-score is \(1.4547859349066146\). Using a z-table or a function that gives the cumulative distribution function of the normal distribution, the probability is approximately \(0.9271357333989796\).
Step 7 ：For part c, the z-scores are \(-0.7273929674533092\) and \(1.2729376930432876\). Using a z-table or a function that gives the cumulative distribution function of the normal distribution, the probability is approximately \(0.6649872931005181\).
Step 8 ：Final Answer: The probability that at least 32 out of 150 18-34 year olds check their social media before getting out of bed is approximately \(\boxed{0.966}\). The probability that no more than 50 do so is approximately \(\boxed{0.927}\). The probability that between 38 and 49 (inclusive) do so is approximately \(\boxed{0.665}\).
