Given sample size n = 50
Sample mean \(\bar { x }\)  = 825
Population mean μ = 900
Population S.D. σ = 110
Null hypotheses: H₀: μ = 900
Alternative hypotheses: H₁: μ ≠ 900
The test statistic, z = \(\frac { \bar { x } -\mu }{ \frac { \sigma }{ \sqrt { n } } } \)
=\(\frac { 825-900 }{ \frac { 110 }{ \sqrt { 50 } } } \) = -4.82
∴ |z| = -4.82
As the significance level is α = 0.05, \(Z_{ \frac { \alpha }{ 2 } }\) = 1.96
Here |z| > \(Z_{ \frac { \alpha }{ 2 } }\) as 4.82 > 1.96
Inference: As |z| > \(Z_{ \frac { \alpha }{ 2 } }\), H₀ is rejected. Hence, we can conclude that mean life time of the population of electric bulbs cannot be taken as 900 hours.

Given sample size n = 64
Sample mean \(\bar { x } \) = 51250
Null hypotheses: H₀: Population mean μ = 50000 Alternative hypotheses: H₁ : μ≠ 50,000
The test statistic, z = \(\frac { \bar { x } -\mu }{ \frac { \sigma }{ \sqrt { n } } } =\frac { 51250-50000 }{ \frac { 2000 }{ \sqrt { 64 } } } \)=5
∴ z = 5
As the significance level is α = 0.05, \(Z_{ \frac { \alpha }{ 2 } }\) = 1.96
Here z > z_α as 5 < 1.96
Inference: As z > z_α, H₀ is rejected. Hence, we can conclude that the sample mean differs significantly from the population mean.

Sample size n = 50
Sample mean \(\bar { x } \) = 75
Sample S.D. s = 10
Standard error (S.E) = \(\frac { s }{ \sqrt { n } } =\frac { 10 }{ \sqrt { 50 } } =\frac { 10 }{ 7.07 } \)
= 1.414
As the significance level is α = 0.005, \(Z_{ \frac { \alpha }{ 2 } }\) = 1.96
∴ 95% confidence limits for the population mean is \(\bar { X } -Z_{ \frac { \alpha }{ 2 } }(S.E)\le \bar { X } -Z_{ \frac { \alpha }{ 2 } }(S.E)\)
⇒ 75 - (1.96)(1.414) ≤ μ ≤ 75 + (1.96)(1.414)
⇒ 75-2.771 ≤ μ ≤ 75 + 2.771
⇒ 72.23 ≤ μ ≤ 77.77
Hence, the 95% confidence interval of the population mean is (72.23, 77.77)

Sample mean \(\bar { x } =\frac { \Sigma x }{ n } \)
= \(\frac { 6.35+6.37+6.36+6.32+6.37 }{ 5 } \)
= \(\frac { 31.75 }{ 5 } \) = 6.35 mm

Sample variance = \(\frac { 1 }{ n-1 } \Sigma (x-\bar { x } )^{ 2 }=\frac { 0.0023 }{ 4 } \)
= 0.00055
n = 0.0055 mm²

Sample size n = 500
Proportion of bad apples in the sample
= \(\frac { 45 }{ 500 } \) = 0.09
p = 0.09
∴ q = 1 - p = 1-0.09 = 0.91
Standard error = \(\sqrt { \frac { pq }{ n } } =\sqrt { \frac { (.09)(.91) }{ 500 } } \)
= 0.0128
As the significance level is α = 0.001, \(Z_{ \frac { \alpha }{ 2 } }\) = 2.58
∴ The confidence limits for the population proportion of bad apples are given by
p-z_α (S.E.) ≤ p ≤ p + z_α (S.E.)
⇒ 0.09 - 2.58 (0.0128) ≤ p ≤ .09 + 2.58 (0.0128)
⇒ 0.09 - 0.033 ≤ p ≤ .09 + 0.033
⇒ 0.057 ≤ p ≤ 0.123
Thus, the bad apples in the consignment lie between (0.057, 0.123).