SOLUTION: Can you please help me solve this Suppose that the times taken for germination for cauliflower seeds are normally distributed with a mean of 7.1 days. Suppose also that exactly 7

Algebra ->  Probability-and-statistics -> SOLUTION: Can you please help me solve this Suppose that the times taken for germination for cauliflower seeds are normally distributed with a mean of 7.1 days. Suppose also that exactly 7      Log On


   

Question 1205463: Can you please help me solve this Suppose that the times taken for germination for cauliflower seeds are normally distributed with a mean of 7.1 days. Suppose also that exactly 70% of the cauliflower seeds germinate in 6.1 days or more. Find the standard deviation of times taken for germination for cauliflower seeds. Carry your intermediate computations to at least four decimal places. Round your answer to at least two decimal places.
Answer by math_tutor2020(3817) About Me  (Show Source):
You can put this solution on YOUR website!

mu = mean = 7.1
sigma = standard deviation = unknown
x = number of days to germinate

"exactly 70% of the cauliflower seeds germinate in 6.1 days or more" translates to the notation P%28x+%3E=+6.1%29+=+0.70
It's practically the same as P%28x+%3E+6.1%29+=+0.70 since the boundary doesn't affect the probability.
The diagram for it will have shading to the right of x = 6.1 underneath the curve. This shaded region has an area of 0.70

We need to find the value of k such that P(z > k) = 0.70
It is equivalent to P(z < k) = 0.30
Note that I'm using z instead of x at this point.

You'll need to use a stats calculator to find the value of k.
You could use a Z table, but it would be fairly inaccurate.

If you have a TI84 calculator or similar then type in:
invNorm(0.30)
The approximate result is -0.5244005
To reach the invNorm function, press the button labeled "2ND" and then press the VARS key.

Many similar calculators can be used as an alternative. Search out "inverse normal calculator".
One result that shows up is
https://onlinestatbook.com/2/calculators/inverse_normal_dist.html
It's from professor David M Lane. The calculator is fairly user friendly and offers a diagram as well.
Two offline alternative routes are the InverseNormal command in GeoGebra or normInv command on a spreadsheet

Anyways, we found that k = -0.5244005 approximately
It indicates P(z < -0.5244005) = 0.30
The area under the standard normal curve to the left of z = -0.5244005 is 0.30
30% of the area under the z curve is to the left of that z score.
70% of the area is to the right of that z score.

The raw score that pairs up with that z value is x = 6.1

The last set of steps look like this:
z = (x - mu)/sigma
z*sigma = x - mu
sigma = (x - mu)/z
sigma = (6.1 - 7.1)/(-0.5244005)
sigma = 1.9069394
sigma = 1.91 when rounding to two decimal places.