In many families both parents work, as a result there is increasing need for day care. Data was collected; and one year in Canada approximately 32% of children aged 0 to 11 years were in day care for at least 20h per week.
What is the probability, in a random poll of 60 children from the ages of 0 to 11 that more than 15 children are in day care at least 20h per week?
p(k)= nCk P^k (1-p)^n-k
were
n= 60
p=0.32
k= 15
q= 1-0.32= 0.68
so:
p(k) = 60C15 x 0.32^15 (0.68) ^60-15
= 0.0583464079
is this correct?
What is the probability in a random poll of 60 children that fewer than 20 are in a day care at least 20h per week?
How would I do this because it says fewer?
Answer is incorrect.
You could answer this exactly by using a combinatorial application as you have tried to do, or you could approximate the answer using the properties of a binominal distribution. I will use the latter. (The formula you gave is the would give the probability of observing EXACTLY 15 children.)
With 60 kids we would expect .32*60=19.2 kids to be in day care. This estimate has an estimated standard deviation of sqrt(n*p*q) = sqrt(60*.32*.68) = 3.613. Now then you want to know the likelyhood of observing more than 15 kids. (I will use a mid-point of 15.5) So, 19.2-15.5 = 3.7. This is (3.7/3.613)=1.02 standard deviations below the mean. Looking in my Cumulative Normal Distribution table, 1.02 corresponds to .8461. ERGO, there is approximately an 85% chance of observing more than 15 children out of a sample of 60 will be in day care.
Follow the same steps for estimating the number fewer than 20. It should be slightly greater than 50%.
To calculate the probability that fewer than 20 children are in day care at least 20 hours per week in a random poll of 60 children, we can use the same approach as before.
We'll estimate the expected number of children in day care as 0.32 * 60 = 19.2. The standard deviation is sqrt(60 * 0.32 * 0.68) = 3.613.
To find the probability of fewer than 20 children, we calculate the z-score, which represents the number of standard deviations below the mean. In this case, it is (20 - 19.2) / 3.613 = 0.221.
Looking up this z-score in a cumulative normal distribution table, we find that it corresponds to approximately 0.5879. This represents the probability of observing fewer than 20 children in day care.
Therefore, in a random poll of 60 children from ages 0 to 11, there is approximately a 58.79% chance that fewer than 20 children will be in day care for at least 20 hours per week.
To find the probability that fewer than 20 children are in day care at least 20 hours per week out of a random poll of 60 children, you can use the same approach as before.
First, calculate the estimated mean number of children in day care: 0.32 * 60 = 19.2 children.
Next, calculate the estimated standard deviation: sqrt(n * p * q) = sqrt(60 * 0.32 * 0.68) ≈ 3.613.
Since you want to find the probability of observing fewer than 20 children, you need to calculate how many standard deviations below the mean 20 is: (20 - 19.2) / 3.613 ≈ 0.221.
Now, use the cumulative normal distribution table or a statistical software to find the probability corresponding to the z-score of 0.221. In this case, you want to find the probability of observing a value less than 0.221, which is approximately 0.5889.
Therefore, there is approximately a 58.89% chance of observing fewer than 20 children out of a random poll of 60 children in day care for at least 20 hours per week.