the average starting salary is Rs 15000per month with a standard deviation of Rs.1000 assuming normal distribution,what is the probability that starting salary will be more than Rs.17000, out of 200 degree holders how many will getRs.14000?

To find the probability that the starting salary will be more than Rs. 17000, you can use the z-score and the standard normal distribution table.

First, let's calculate the z-score. The formula to calculate the z-score is:

z = (X - μ) / σ

where:
X = the value we want to find the probability for (Rs. 17000 in this case)
μ = the mean (average starting salary) = Rs. 15000
σ = the standard deviation = Rs. 1000

Plugging in the values, we get:

z = (17000 - 15000) / 1000
z = 2

Now, we can use the standard normal distribution table to find the probability corresponding to the z-score of 2. The standard normal distribution table gives the probability of finding a z-score less than a given value, but we want the probability of finding a z-score greater than 2.

Since the distribution is symmetric, we can find the probability of finding a z-score greater than 2 by subtracting the probability of finding a z-score less than 2 from 1.

Using the standard normal distribution table, the probability of finding a z-score less than 2 is approximately 0.9772. Therefore, the probability of finding a z-score greater than 2 is:

1 - 0.9772 = 0.0228

So, the probability that the starting salary will be more than Rs. 17000 is approximately 0.0228 or 2.28%.

Next, to find out how many out of 200 degree holders will get Rs. 14000, we can use the concept of the normal distribution and the z-score.

The z-score formula remains the same:

z = (X - μ) / σ

where:
X = the value we want to find the number of occurrences for (Rs. 14000 in this case)
μ = the mean (average starting salary) = Rs. 15000
σ = the standard deviation = Rs. 1000

Calculating the z-score for Rs. 14000:

z = (14000 - 15000) / 1000
z = -1

To find the number of occurrences for a given z-score, we need to convert the z-score into a probability and then multiply it by the total number of occurrences (200 in this case).

Using the standard normal distribution table, the probability of finding a z-score less than -1 is approximately 0.1587.

So, the number of degree holders who will get Rs. 14000 is:

0.1587 * 200 = 31.74

Since we cannot have fractional degree holders, the number of degree holders who will get Rs. 14000 would be approximately 32.