Answer:
Yes, the temperature is within the confidence interval (37.40, 37.60)
Explanation:
Mean=37.48, sd=0.6, n=100, df=n-1=100-1=99, t value corresponding to 99 degrees of freedom and 90% confidence interval is 1.6602
CI = (mean + or - t×sd/√n)
CI = (37.48 + 1.6602×0.6/√100) = 37.48 + 0.0996 = 37.60
CI = (37.48 - 1.6602×0.6/√100) = 37.48 - 0.0996 = 37.40
CI is (37.40, 37.60)