I solved following question, but I'm not sure whether I've taken the right approach Any help is highly appreciated!

Question

Question:

An article in the Journal of Agricultural Science [“The Use of Residual Maximum Likelihood to Model Grain Quality Characteristics of Wheat with Variety, Climatic and Nitrogen Fertilizer Effects” (1997, Vol. 128, pp. 135–142)] investigated means of wheat grain crude protein content (CP) and Hagberg falling number (HFN) surveyed in the UK. The analysis used a variety of nitrogen fertilizer applications (kg N/ha), temperature (ºC), and total monthly rainfall (mm). The data shown below describe temperatures for wheat grown at Harper Adams Agricultural College between 1982 and 1993. The temperatures measured in June were obtained as follows: 15.2 14.2 14.0 12.2 14.4 12.5 14.3 14.2 13.5 11.8 15.2 Assume that the standard deviation is known to be σ = 0.5.

part (a) Suppose that we wanted to be 95% confident that the error in estimating the mean temperature is less than 2 degrees Celsius. What sample size should be used?

part (b) Suppose that we wanted the total width of the two-sided confidence interval on mean temperature to be 1.5 degrees Celsius at 95% confidence. What sample size should be used?

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My attempt so far:

part (a) my attempt:

So we know, E = ( z(alpha/2) * standard deviation )/(sqrt(sample size)) and E < 2(given)

From here we get, ( z(alpha/2) * standard deviation )/(sqrt(sample size)) < 2

which gives, n > ( z(alpha/2) * standard deviation / E )^2

==> n > (1.96*0.5/2)^2

==> n > (1.96*/4)^2

==> n > (0.49)^2 = 0.2401

==> n > 0.2401 and by rounding up n, we get,
n > 1(approximately) for part (a)

IS THIS CORRECT????

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part (b) - my attempt

So, the width of interval = [ sample mean + ( z(alpha/2) * standard deviation )/(sqrt(sample size)) ] - [ sample mean - ( z(alpha/2) * standard deviation )/(sqrt(sample size)) ]

which gives, width of the interval = 2*( z(alpha/2) * standard deviation )/(sqrt(sample size))

But, it is given that the width of the interval = 1.5 which gives,

n = 4*[( z(alpha/2) * standard deviation )^2]/ (width of interval^2)

Simplifying, n = 4*[1.96*0.5/1.5]^2 = 4*(1/96/3)^2 = 1.7074

Again by rounding up n, we get,

n = 2(approximately) for part (b)

IS THIS CORRECT??