Let X be the iq score of adults. X is normally distributed with mean of 100 and standard deviation of 20.
If all iq scores are converted to z -scores then we will get z scores. And the z score is standard normal variable with mean=0 and standard deviation =1
Hence the mean and standard deviation of of all z scores will be 0 and 1 respectively.
Answer:
More than 50% would germinate
Step-by-step explanation:
Given that in the production of a plant, a treatment is being evaluated to germinate seed. From a total of 60 seed it was observed that 37 of them germinated
Let us check whether more than 50% will germinate using hypothesis test

(right tailed test)
Sample proportion p =
p difference = 0.117
Test statistic Z = p difference/std error = 1.864
p value =0.0312
Since p value <0.05 our significance level of 5% we reject null hypothesis
It is possible to claim that most of the seed will germinate (i.e. more than 50%)
A=(1/2)(r^2)(sin[360/n])(n)
A=(1/2)(8^2)(sin[360/15])(15)
A=(1/2)(64)(sin[24°])(15)
A=(1/2)(64)(0.40673664)(15)
A= 195.2335872
Step-by-step explanation:
3x-3(20)=12
3x=72
x=24.
24 stickers on each sheet