Normal distirubtion question help agaiiinnn?

ne help with steps on how to do these ques wud b much appreciated :) thnnnnx


1) a variety of hollyhock grows to great heights. assuming a normal distribution of heights , find the mean and standard deviation, given that the 30th and 70th percentiles are 1.83m and 2.31 m respectively. (answer=2.07m, 0.458m)





2) the lengths of time spent each night by ulrika in watching tv are observations from a normal distribution, with a mean of 190 minutes and a standard deviation of 27 minutes. Determine to the nearest minute, the length of time spent watching which Ulrika only exceeds on one night in 200 on average. (answer=260min)

Normal distirubtion question help agaiiinnn?
1. we know that P(Z %26lt; z1=(1.83 - μ)/σ) = 0.3


while P(Z %26lt; z2=(2.31 - μ)/σ) = 0.7





and by the way... 0.3 %26amp; 0.7 are of the same distance from the mean... thus the mean is the average of 1.83 %26amp; 2.31 which is 2.07 = μ





then P(Z %26lt; z1=(1.83 - 2.07)/σ) = 0.3





z1 = -0.524 = (1.83 - 2.07)/σ


{using a table or a calculator...}





σ = -0.24/(-0.524) = 0.458








§





2. we are looking for x1 such that





z1 = (x1 - 190)/27





such that P(Z %26lt; z1) = 199/200 = 0.995





z1 = 2.58


x1 = 190 + 27*2.58 = 260
Reply:1) P( x %26lt; 1.83) =0.30


P( x %26lt; 2.31) =0.70


From the Normal table, the 30th percentile is -0.52 and the 70th percentile is 0.52.


(1.83-mean)/sd = -0.52


-0.52*sd+mean=1.83


0.52*sd+mean=2.31 --- (1)


-1.04sd=-0.48


sd=0.48/1.04=0.4615 (close enough to 0.458)


plug this sd into (1) and solve for mean


0.52(0.458)+mean=2.31


mean = 2.31 -0.52(0.458) =2.07184