Asked by ken
Relative humidities (represented by RH below) of less than 51% cause problems in growing crops. The Saskatoon region is a prime wheat-growing area. The mean RH in Saskatoon is 59% in July, with a standard deviation of 8%. The RH for the days in July forms a normal distribution, on how many days in July is it likely to be dry enough to threaten wheat crops in the area?
(a) What percent of days in July is it likely to be dry enough to threaten wheat crops? In other words, in what percent of the days is the RH below 51%?
(b) On how many days in July is it likely to be dry enough to threaten wheat crops in the area? Of course everyone knows that July has 31 days!
(a) What percent of days in July is it likely to be dry enough to threaten wheat crops? In other words, in what percent of the days is the RH below 51%?
(b) On how many days in July is it likely to be dry enough to threaten wheat crops in the area? Of course everyone knows that July has 31 days!
Answers
Answered by
PsyDAG
Z = (x - μ)/SD
Z = (51 - 59)/8
a) Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion below that Z score.
b) Multiply the above proportion by 31.
Z = (51 - 59)/8
a) Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion below that Z score.
b) Multiply the above proportion by 31.
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