Answer :
The lower limit for the 95%confidence interval for the given mean and standard deviation is equal to 39.03.
As given in the question,
Sample size 'n' = 60
Mean 'μ' = $40.80
Standard deviation 'σ' = $7.00
Using z-score table of normal distribution
z-value for 95%confidence interval = 1.96
Formula used
Confidence interval
= μ ± ( z-value × σ ) /√n
= 40.80 ± (1.96 × 7)/√60
= 40.80 ± (13.72 /7.75)
= 40.80 ±1.770
Lower limit of the confidence interval is equal to :
40.80 - 1.770 = 39.03
Upper limit of the confidence interval is equal to :
40.80 + 1.770 = 42.57
Therefore, the lower limit for the 95% confidence interval is equal to 39.03.
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