A farmer finds that the weights of sheep on his farm have a normal distribution with mean 66.4 kg
and standard deviation 5.6 kg.
250 sheep are chosen at random. Estimate the number of sheep which have a weight of between
70 kg and 72.5 kg.
You can play around with Z table stuff at
davidmlane.com/hyperstat/z_table.html
70 kg is [(70 - 66.4) / 5.6] s.d. above the mean
72.5 kg is [(72.5 - 66.4) / 5.6] s.d. above the mean
calculate the two z-scores (s.d. above the mean)
... then use a table to find the portion of the population between the scores
I need to know how to do.I tried it and got 0.13128, but the answer is 0.1221.
Do we need continuity correction also?If yes then why?
there is a z-score calculator at calculator.net
it gives greater precision than the tables
... no need for interpolation from table values
To estimate the number of sheep with weights between 70 kg and 72.5 kg, we can use the concept of standard deviation and the properties of a normal distribution. Here's how you can calculate it:
1. Calculate the z-scores for the given weights using the formula:
z = (x - μ) / σ
Where:
- x is the weight you want to find the z-score for (either 70 kg or 72.5 kg)
- μ is the mean of the distribution (66.4 kg)
- σ is the standard deviation of the distribution (5.6 kg)
For 70 kg:
z1 = (70 - 66.4) / 5.6
For 72.5 kg:
z2 = (72.5 - 66.4) / 5.6
2. Look up the corresponding probabilities for the z-scores in the standard normal distribution table. The table gives you the area under the curve to the left of each z-score.
Let's assume the probability for z1 is P1 and the probability for z2 is P2.
3. Calculate the estimated number of sheep in the given weight range using the formula:
Estimated number = (P2 - P1) * total number of sheep
Where:
- P2 is the probability corresponding to z2
- P1 is the probability corresponding to z1
- Total number of sheep is given as 250.
By performing these steps, you will be able to estimate the number of sheep with weights between 70 kg and 72.5 kg based on the given mean and standard deviation.