Lower quartile=39, Upper quartile=63.

Measurements of wind speed on a certain island were taken over a period of one year. A box-andwhisker
plot of the data obtained is displayed above, and the values of the quartiles are as shown.
It is suggested that wind speed can be modelled approximately by a normal distribution with mean
μ kmh−1 and standard deviation s kmh−1.
(i) Estimate the value of μ. [1]
(ii) Estimate the value of s.

Lol. No

(i) Well, if the lower quartile is 39 and the upper quartile is 63, then the median would be smack dab in the middle, at 51. So, let's estimate that the mean μ would also be around 51 km/h. However, keep in mind that this is just an estimate!

(ii) Now, to estimate the value of s (the standard deviation), we can use the interquartile range (IQR), which is the difference between the upper quartile and the lower quartile. In this case, the IQR would be 63 - 39 = 24 km/h.

Since the normal distribution captures about 68% of the data within one standard deviation, we can guess that s would be approximately equal to IQR/2. Therefore, we can estimate that s is around 12 km/h. But hey, remember, this is just an educated guess!

To estimate the value of μ (the mean), we can consider the midpoint of the quartiles, which is the median.

The median is the value that separates the lower half of the data from the upper half. In a normal distribution, the median is equal to the mean.

Given that the lower quartile is 39 and the upper quartile is 63, the median would be the average of these two values:

Median = (Lower Quartile + Upper Quartile) / 2
= (39 + 63) / 2
= 102 / 2
= 51 km/h

Therefore, we can estimate that the value of μ (the mean) is approximately 51 km/h.

To estimate the value of s (the standard deviation), we can use the interquartile range (IQR). The IQR is the range between the upper and lower quartiles and represents the spread of the data.

IQR = Upper Quartile - Lower Quartile
= 63 - 39
= 24 km/h

In a normal distribution, approximately 50% of the data falls within ±1 standard deviation from the mean. Since the IQR represents roughly 50% of the data, we can estimate the standard deviation as:

Standard Deviation (s) ≈ IQR / 1.34896

where 1.34896 is an approximation of the standard deviation for a normal distribution.

s ≈ 24 / 1.34896
≈ 17.8 km/h

Therefore, we can estimate that the value of s (the standard deviation) is approximately 17.8 km/h.

Ans are (i)51 and (ii)17.8. pls help

X~N (51,s^2)

P(X<39)=0.25
solve to find s
Ans: 17.804