The approximate value of P(z ≤ -1.25) will be option (c) 11%
Normal distribution
A probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean.
What are the steps to solve this problem?
The steps are as follow:
- Given P(z ≤ -1.25)
- In the given table on the first row and column represents values of z
- For -1.25 we have to see value of z at -1.2 from column and 0.05 from row and where they meet will be our value of P(z ≤ -1.25)
- Here the value we get is 0.1056
- If we round off this value will get 0.11 so it will be 11%
Hence for P(z ≤ -1.25) from standard normal distribution the approximate value will be 11% which is option (c).
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