00:01
Let us represent x is the weight of a female cat and it is given that x follows the normal distribution with the mean mu equal to 4 .1 kg and with the standard deviation sigma equal to 0 .6 kg.
00:17
In part a we have to find the probability that the cat weight is between 3 .7 and 4 .4 kg and that is calculated as probability that to standardize the value.
00:30
3 .7 minus the mean value of 4 .1 divided by the standard deviation value of 0 .6, which is less than x minus mu by sigma, which is less than 4 .4 minus 4 .1 divided by 0 .6.
00:45
Now simplifying this gives probability that minus 0 .67, which is less than z, which is less than 0 .5.
00:54
Since z is a symmetric distribution, this probability can be written as probability that, z is less than 0 .5 minus probability that z is less than minus 0 .67.
01:06
Now using the standard number distribution table, this probabilities are 0 .6915 minus 0 .2514.
01:15
And simplifying this gives the probability that 0 .4401.
01:20
Thus, this is the required probability.
01:23
In part b, it is given that a certain cat x is 0 .5 standard, above the mean of the distribution and it is given that this cat as a z value of 0 .5.
01:41
Now we have to find the proportion of cats that are greater than the given conditions.
01:48
That is, we have to find the probability that the standard normal variant is greater than the given value of 0 .5.
01:57
From the standard normal distribution table, this probability is 0 .6915.
02:05
In path c, we have to find the 80th percentile of the data.
02:12
Now to find the z value for this 80th percentile, the excel formula of equal to norm...