Workload, pressure to complete tasks, and time urgency associated with = Challenge Stressors
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Answer:
The probability that the mean receipt for dinner is between $24.50 and $25.50 is
P(24.50≤ x ≤ 25.50)= 0.8324
Step-by-step explanation:
<u><em>Step(i)</em></u>:-
Given mean of Population(μ) = $25
Given standard deviation of the Population (σ) = 2.30
Sample size 'n' = 40
Let 'X' be the random variable in normal distribution
Given X₁ = 24.50
Given X₂ = 25.50
<u><em>Step(ii):-</em></u>
The probability that the mean receipt for dinner is between $24.50 and $25.50
P(x₁≤ x ≤ x₂) = P(Z₁≤ z ≤ Z₂) = A(Z₂) + A(z₁)
P(24.50≤ x ≤ 25.50) = P(-1.375≤ z ≤ 1.375) = A(1.375) + A(1.375)
P(24.50≤ x ≤ 25.50)= 2 A( 1.375)
= 2 × 0.4162 ( from normal table)
= 0.8324
<u><em>Final answe</em></u>r:-
The probability that the mean receipt for dinner is between $24.50 and $25.50 is 0.8324
Answer:
Step-by-step explanation:
sqrt of 7-11=2i
simplify to
then sqrt of 9=3
then we simplify again to
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